Question 3515 Marks
Evaluate the following integrals:
$\int\sin^{-1}\sqrt{\frac{\text{x}}{\text{a+x}}}\text{dx}$
AnswerLet $\text{I}=\int\sin^{-1}\sqrt{\frac{\text{x}}{\text{a+x}}}\text{dx}$
Let $\text{x}=\text{a}\tan^2\theta$
$\text{dx}=2\text{a}\tan\theta\sec^2\theta\text{d}\theta$
$\text{I}=\int\Big(\sin^{-1}\sqrt{\frac{\text{a}\tan^2\theta}{\text{a+a}\tan^2\theta}}\Big)(2\text{a}\tan\theta\sec^2\theta)\text{d}\theta$
$=\int\Big(\sin^{-1}\sqrt{\frac{\tan^2\theta}{\sec^2\theta}}\Big)(2\text{a}\tan\theta\sec^2\theta)\text{d}\theta$
$=\int\sin^{-1}(\sin\theta)(2\text{a}\tan\theta\sec^2\theta)\text{d}\theta$
$=\int2\theta\text{a}\tan\theta\sec^2\theta\text{d}\theta$
$=2\text{a}\int\theta(\tan\theta\sec^2\theta)\text{d}\theta$
$=2\text{a}\big[\theta\int\tan\theta\sec^2\theta\text{d}\theta-\int(\int\tan\theta\sec^2\theta\text{d}\theta)\text{d}\theta\Big]$
$=2\text{a}\Big[\theta\frac{\tan^2\theta}{2}-\int\frac{\tan^2\theta}{2}\text{d}\theta\Big]$
$=\text{a}\theta\tan^2\theta-\frac{2\text{a}}{2}\int(\sec^2\theta-1)\text{d}\theta$
$=\text{a}\theta\tan^2\theta-\text{a}\tan\theta+\text{a}\theta+\text{C}$
$=\text{a}\Big(\tan^{-1}\sqrt{\frac{\text{x}}{\text{a}}}\Big)\frac{\text{x}}{\text{a}}-\text{a}\sqrt{\frac{\text{x}}{\text{a}}}+\text{a}\tan^{-1}\sqrt{\frac{\text{x}}{\text{a}}}+\text{C}$
$\text{I}=\text{x}\tan^{-1}\sqrt{\frac{\text{x}}{\text{a}}}-\sqrt{\text{ax}}+\text{a}\tan^{-1}\sqrt{\frac{\text{x}}{\text{a}}}+\text{C}$
View full question & answer→Question 3525 Marks
Evaluate the following integrals:
$\int\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
AnswerLet $\text{I =}\int\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
Putting $\text{x}=\cos\theta$
$\Rightarrow\text{dx}=-\sin\theta\text{d}\theta$
$\&\theta=\cos^{-1}\text{x}$
$\therefore\text{I}=\int\tan^{-1}\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}(-\sin\theta)\text{d}\theta$
$=\int\tan^{-1}\sqrt{\frac{2\sin^2\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}}(-\sin\theta)\text{d}\theta$
$=\int\tan^{-1}\Big(\tan\frac{\theta}{2}\Big)(-\sin\theta)\text{d}\theta$
$=-\frac{1}{2}\int\theta\sin\theta\text{d}\theta$
$=-\frac{1}{2}\Big[\theta\int\sin\theta\text{d}\theta-\int\Big\{\Big(\frac{\text{d}}{\text{d}\theta}\theta\Big)\int\sin\theta\text{d}\theta\Big\}\text{d}\theta\Big]$
$=-\frac{1}{2}\big[\theta(-\cos\theta)-\int1.(-\cos\theta)\text{d}\theta\big]$
$=-\frac{1}{2}\big[-\theta\cos\theta+\sin\theta\big]+\text{C}$
$=-\frac{1}2{}\Big[-\theta.\cos\theta+\sqrt{1-\cos^2\theta}\Big]+\text{C}$
$=-\frac{1}{2}\Big[-\cos^{-1}\text{x.x}+\sqrt{1-\text{x}^2}\Big]+\text{C}$ $\big[\because\theta=\cos^{-1}\text{x}\big]$
$=\frac{\text{x}\cos^{-1}\text{x}}{2}-\frac{\sqrt{1-\text{x}^2}}{2}+\text{C}$
View full question & answer→Question 3535 Marks
Evaluvate the following intregals:
$\int\frac{1}{1-\cot\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{1-\cot\text{x}}\text{ dx}$
$=\int\frac{1}{1-\frac{\cos\text{x}}{\sin\text{x}}}\ \text{dx}$
$=\int\frac{\sin\text{x}}{\sin\text{x}-\cos\text{x}}\ \text{dx}$
Let $\sin\text{x}=\lambda\frac{\text{d}}{\text{dx}}(\sin\text{x}-\cos\text{x})+\mu(\sin\text{x}-\cos\text{x})+\text{v}$
$\sin\text{x}=\lambda\frac{\text{d}}{\text{dx}}(\cos\text{x}+\sin\text{x})+\mu(\sin\text{x}-\cos\text{x})+\text{v}$
$\sin\text{x}=\cos(\lambda-\mu)+\sin\text{x}(\lambda+\mu)+\text{v}$
Compairing the cooefficients of $\sin\text{x}\ \&\cos\text{x}$ on the both the sides,
$\lambda+\mu=1\ ...(1)$
$\lambda-\mu=1\ ...(2)$
$\text{v}=0\ ...(3)$
Equation (1), (2), (3) gives
$\lambda=\frac{1}{2},\mu=\frac{1}{2},\text{v}=0$
$\text{I}=\int\frac{\frac{1}{2}(\cos\text{x}+\sin\text{x})+\frac{1}{2}(\sin\text{x}-\cos\text{x}_)}{(\sin\text{x}-\cos\text{x})}\ \text{dx}$
$=\frac{1}{2}\int\frac{(\cos\text{x}+\sin\text{x})}{(\sin\text{x}-\cos\text{x})}\ \text{dx}+\frac{1}{2}\int\ \text{dx}$
$\text{I}=\frac{1}{2}\log|\sin\text{x}-\cos\text{x}|+\frac{1}{2}\text{x}+\text{c}$
View full question & answer→Question 3545 Marks
$\int\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}\text{dx}$
$=\int\frac{\text{x}^2}{(\text{x}^2-4)(\text{x}^2+3)}\text{dx}$
Now, $\frac{\text{x}^2}{(\text{x}^2-4)(\text{x}^2+3)}=\frac{\text{A}}{\text{x}^2-4}+\frac{\text{B}}{\text{x}^2+3}$
Let $\text{x}^2=\text{t}$
$\Rightarrow\ \frac{\text{t}}{(\text{t}-4)(\text{t}+3)}=\frac{\text{A}}{\text{t}-4}+\frac{\text{B}}{\text{t}+3}$
$\Rightarrow \text{t}=\text{A}(\text{t}+3)+\text{B}(\text{t}-4)$
On comparting the coefficient of t on both sides, we get
$\text{A}+\text{B}=1\ \ \dots(\text{i})$
Comparing constant term, we get
$3\text{A}-4\text{B}-0\ \ \dots(\text{ii})$
Solving (i) and (ii), we get
$\therefore\ \text{A}=\frac{4}{7}$ and $\text{B}=\frac{3}{7}$
$\frac{\text{x}^2}{(\text{x}^2-4)(\text{x}^2+3)}=\frac{4}{7(\text{x}^2-4)}+\frac{3}{7(\text{x}^2+3)}$
$\therefore\ \text{I}=\frac{4}{7}\int\frac{1}{\text{x}^2-2^2}\text{dx}+\frac{3}{7}\int\frac{1}{\text{x}^2+(\sqrt{3})^2}\text{dx}$
$=\frac{4}{7}\cdot\frac{1}{2.2}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{3}{7}\cdot\frac{1}{\sqrt{3}}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$
$=\frac{1}{7}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{\sqrt{3}}{7}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$
View full question & answer→Question 3555 Marks
Evaluate the following integrals:
$\int\frac{1}{\cos\text{x}(5-4\sin\text{x})}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sin\text{x}\sin2\text{x}}\ \text{dx}$
$\int\frac{\sin\text{x dx}}{\sin^2\text{x}+2\sin\text{x}\cos\text{x}}$
$=\int\frac{\sin\text{x dx}}{(1-\cos^2\text{x})+2(1-\cos^2\text{x})\cos\text{x}}$
Let $\cos\text{x}=\text{t}$
$\Rightarrow-\sin\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{(\text{t}^2-1)+2(\text{t}^2-1)\text{t}}$
Let $\frac{1}{(\text{t}^2-1)(3+2\text{t})}=\frac{\text{A}}{\text{t}-1}+\frac{\text{B}}{\text{t}+1}+\frac{\text{C}}{1+2\text{t}}$
$\Rightarrow 1 = A(t + 1)(1 + 2t) + B (t - 1)(1 + 2t) + C (t^2 - 1)$
Put $t = 1$
$\Rightarrow 1 = 6A $
$\Rightarrow $$\text{A}=\frac{1}{6}$
Put $t = -1$
$\Rightarrow 1 = 2B$
$\Rightarrow $ $\text{B}=-\frac{1}{2}$
Put $\text{t}=-\frac{1}{2}$
$\Rightarrow1=-\frac{3}{4}\text{C}\Rightarrow\text{C}=-\frac{4}{3}$
thus,
$\text{I}=\frac{1}{6}\int\frac{\text{dt}}{\text{t}-1}+\frac{1}{2}\int\frac{\text{dt}}{\text{t}+1}-\frac{4}{3}\int\frac{\text{dt}}{1+2\text{t}}$
$=\frac{1}{6}\log|\text{t}-1|+\frac{1}{2}\log|\text{t}+1|-\frac{2}{3}\log|1+2\text{t}|+\text{}C$
Hence,
$\text{I}=\frac{1}{6}\log|\cos\text{x}-1|+\frac{1}{2}\log|\cos\text{x}+1|+\frac{2}{3}\log|1+2\cos\text{x}|+\text{C}$
View full question & answer→Question 3565 Marks
Evaluate the following intregals:
$\int\frac{1}{(\sin\text{x}-2\cos\text{x})(2\sin\text{x}-\cos\text{x})}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{(\sin\text{x}-2\cos\text{x})(2\sin\text{x}-\cos\text{x})}\ \text{dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\Rightarrow\text{I}=\int\frac{\sec^2\text{x}}{\Big(\frac{\sin\text{x}-2\cos\text{x}}{\cos\text{x}}\Big)\times\Big(\frac{2\sin\text{x}+\cos\text{x}}{\cos\text{x}}\Big)}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{(\tan^2\text{x}-2)(2\tan\text{x}+1)}\ \text{dx}$
Let $\tan\text{x}=\text{t}$
$\Rightarrow\sec^2\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{(\text{t}-2)(2\text{t}-1 0)}$
$=\int\frac{\text{dt}}{2\text{t}^2+\text{t}-4\text{t}-2}$
$=\int\frac{\text{dt}}{2\text{t}^2-3\text{t}-2}$
$=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2-\frac{3}{2}\text{t}-1}$
$=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2-\frac{3}{2}\text{t}+\Big(\frac{3}{4}\Big)^2-\Big(\frac{3}{4}\Big)^2-1}$
$=\frac{1}{2}\int\frac{\text{dt}}{\Big(\text{t}-\frac{3}{4}\Big)-\frac{9}{16}-1}$
$=\frac{1}{2}\int\frac{\text{dt}}{\Big(\text{t}-\frac{3}{4}\Big)-\Big(\frac{5}{4}\Big)^2}$
$=\frac{1}{2}\times\frac{1}{2\times\frac{5}{4}}\log\Bigg|\frac{\text{t}-\frac{3}{4}-\frac{5}{4}}{\text{t}-\frac{3}{4}+\frac{5}{4}}\Bigg|+\text{C}$
$=\frac{1}{5}\ln\Bigg|\frac{\text{t}-2}{\text{t}+\frac{1}{2}}\Bigg|+\text{C}$
$=\frac{1}{5}\ln\Big|\frac{\text{t}-2}{2\text{t}+1}\Big|+\frac{1}{5}\int(2)+\text{C}$
$=\frac{1}{5}\ln\Big|\frac{\text{t}-2}{2\text{t}+1}\Big|+\text{C}$ where $\text{C}=\text{C}+\frac{1}{5}\int(2)$
$=\frac{1}{5}\ln\Big|\frac{\tan\text{x}-2}{2\tan\text{x}+1}\Big|+\text{C}$
View full question & answer→Question 3575 Marks
Evaluate the following integrals:$\int\frac{(3\sin\text{x}-2)\cos\text{x}}{13-\cos^2\text{x}-7\sin\text{x}}\text{ dx}$
Answer$\text{I}=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{13-\cos^2\text{x}-7\sin\text{x}}\text{ dx}$ $=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{13-(1-\sin^2\text{x})-7\sin\text{x}}\text{ dx}$ $\big(\because\ \cos^2\text{x}=1-\sin^2\text{x}\big)$ $=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{\sin^2\text{x}-7\sin\text{x}+12}\text{ dx}$ $=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{\sin^2\text{x}-4\sin\text{x}-3\sin\text{x}+12}\text{ dx}$ $=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{\sin\text{x}(\sin\text{x}-4)-3(\sin\text{x}-4)}\text{dx}$ $=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{(\sin\text{x}-3)(\sin\text{x}-4)}\text{ dx}$Let $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{(3\text{t}-2)}{(\text{t}-3)(\text{t}-4)}\text{ dt}$
Using partial fraction, we get $\frac{(3\text{t}-2)}{(\text{t}-3)(\text{t}-4)}=\frac{\text{A}}{(\text{t}-3)}+\frac{\text{B}}{(\text{t}-4)}$ $=\frac{\text{A}(\text{t}-4)+\text{B}(\text{t}-3)}{(\text{t}-3)(\text{t}-4)}$ $\Rightarrow3\text{t}-2=(\text{A}+\text{B})\text{t}-4\text{A}-3\text{B}$ Comparing coefficients, we get A = -7 and B = 10 So, $\text{I}=-7\int\frac{1}{(\text{t}-3)}\text{ dt}+10\int\frac{1}{(\text{t}-4)}\text{ dt}$$\Rightarrow\text{I}=-7\ln|\text{t}-3|+10\ln|\text{t}-4|+\text{C}$
$\therefore\ \text{I}=-7\ln|\sin\text{x}-3|+10\ln|\sin\text{x}-4|+\text{C}$
View full question & answer→Question 3585 Marks
Evaluvate the following intregals:
$\int\frac{2\tan\text{x}+3}{3\tan\text{x}+4}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{2\tan\text{x}+3}{3\tan\text{x}+4}\ \text{dx}$
$=\int\frac{2\sin\text{x}+3\cos\text{x}}{3\sin\text{x}+4\cos\text{x}}\ \text{dx}$
Let $2\sin\text{x}+3\cos\text{x}=\lambda\frac{\text{d}}{\text{dx}}(3\sin\text{x}+4\cos\text{x})+\mu(3\sin\text{x}+4\cos\text{x})+\text{v}$
$2\sin\text{x}+3\cos\text{x}=\lambda(3\cos\text{x}-4\sin\text{x})+\mu(3\sin\text{x}+4\cos\text{x})+\text{v}$
$2\sin\text{x}+3\cos\text{x}=(3\lambda+4\mu)\cos\text{x}+(-4\lambda+3\mu)\sin\text{x}+\text{v}$
Comparing the coefficient of $\sin\text{x}\ \&\cos\text{x}$ on the both the sides,
$3\lambda+4\mu=3\ \dots\dots(1)$
$-4\lambda+3\mu=2\ \dots\dots(2)$
Solving the equation (1), (2) and (3),
$\mu=\frac{18}{25}$
$\lambda=\frac{1}{25}$
$\text{V}=0$
$\text{I}=\frac{1}{25}\int\frac{(3\cos\text{x}-4\sin\text{x})}{(3\sin\text{x}+4\cos\text{x})}\ \text{dx}+\frac{18}{25}\int\text{dx}$
$\text{I}=\frac{18}{25}\text{x}+\frac{1}{25}\log|3\sin\text{x}+4\cos\text{x}|+\text{C}$
View full question & answer→Question 3595 Marks
Evaluate the following intregals:
$\int\frac{1}{2+\sin\text{x}+\cos\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{2+\sin\text{x}+\cos\text{x}}\text{dx}$
Putting $\sin\text{x}=\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}},\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$\text{I}=-\int\frac{1}{2+\begin{pmatrix}\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\end{pmatrix}+\begin{pmatrix}\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\end{pmatrix}}\text{dx}$
$=\int\frac{\begin{pmatrix}1+\tan^2\frac{\text{x}}{2}\end{pmatrix}}{2+2\tan^2\frac{\text{x}}{2}+2\tan\frac{\text{x}}{2}+1-\tan^2\frac{\text{x}}{2}}\text{dx}$
$=\int\frac{\begin{pmatrix}\sec^2\frac{\text{x}}{2}\end{pmatrix}}{\tan^2\frac{\text{x}}{2}+\tan\frac{\text{x}}{2}+3}\text{dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$
$\text{I}=\int\frac{2\text{dt}}{\text{t}^2+2\text{t}+3}$
$\text{I}=2\int\frac{\text{dt}}{\text{t}^2+2\text{t}+1-1+3}$
$\text{I}=2\int\frac{\text{dt}}{(\text{t}+1)^2+(\sqrt2)^2}$
$=\frac{2}{\sqrt2}\tan^{-1}\Big(\frac{\text{t}+1}{\sqrt2}\big)+\text{C}$
$\text{I}=\sqrt{2}\tan^{-1}\Big(\frac{\tan\frac{\text{x}}{2}+1}{\sqrt2}\Big)+\text{C}$
View full question & answer→Question 3605 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}\ \text{dx}$We express
$\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}=\frac{\text{x}^2}{\text{x}^4-4\text{x}^2+3\text{x}^2-12}$
$=\frac{\text{x}^2}{(\text{x}^2-4)(\text{x}^2+3)}$
$=\frac{\text{A}}{\text{x}^2-4}+\frac{\text{B}}{\text{x}^2+3}$
$\Rightarrow\text{x}^2=\text{A}(\text{x}^2+3)+\text{B}(\text{x}^2-4)$
Equating the coefficients of $x^2$ and constant, we get
1 = A + B and 0 = 3A - 4B or
$\text{A}=\frac{4}{7}$ and $\text{B}=\frac{3}{7}$
$\therefore\text{I}=\int\bigg(\frac{\frac{4}{7}}{\text{x}^2-4}+\frac{\frac{3}{7}}{\text{x}^2+3}\bigg)\text{dx}$
$=\frac{4}{7}\int\frac{1}{\text{x}^2-4}\ \text{dx}+\frac{3}{7}\int\frac{1}{\text{x}^2+3}\text{dx}$
$=\frac{4}{7}\times\frac{1}{4}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{\sqrt{3}}{7}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$
$=\frac{1}{7}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{\sqrt{3}}{7}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$
Hence, $\int\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}\ \text{dx}=\frac{1}{7}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{\sqrt{3}}{7}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$
View full question & answer→Question 3615 Marks
Evaluate the following integrals:$\int\sin^{-1}\sqrt{\text{x}}\text{dx}$
Answer$\int\sin^{-1}\sqrt{\text{x}}\text{dx}$
Let $\text{x} = \text{t}^{2}\ \ \ [\therefore\text{dx = 2tdt}]$
$\int\sin^{-1}\sqrt{\text{x}}\text{dx}=\int\sin^{-1}\sqrt{\text{t}^2}2\text{tdt}=\int\sin^{-1}\text{t}2\text{tdt}$
$=\sin^{-1}\text{t}\int2\text{tdt}-\Big(\int\frac{\text{d}\sin^{-1}\text{t}}{\text{dt}}\big(\int2\text{tdt}\big)\text{dt}\Big)$
$=\sin^{-1}\text{t}(\text{t}^2)-\int\frac{1}{\sqrt{1-\text{t}^2}}(\text{t}^2)\text{dt}$
Lets solve $\int\frac{1}{\sqrt{1-\text{t}^2}}(\text{t}^2)\text{dt}$
$\int\frac{1}{\sqrt{1-\text{t}^2}}(\text{t}^2)\text{dt}=\int\frac{\text{t}^2-1+1}{\sqrt{1-\text{t}^2}}\text{dt}=\int\frac{\text{t}^2-1}{\sqrt{1-\text{t}^2}}\text{dt}+\int\frac{1}{\sqrt{1-\text{t}^2}}\text{dt}$
We know that, value of $\int\frac{1}{\sqrt{1-\text{t}^2}}\text{dt}=\sin^{-1}\text{t}$
Remaining integral to evalute is $\int\frac{\text{t}^2-1}{\sqrt{1-\text{t}^2}}\text{dt}=\int-\sqrt{1-\text{t}^2}\text{dt}$
sub $\text{t}=\sin\text{u},\text{dt}=\cos\text{u du}$
$\int-\sqrt{1-\text{t}^2}\text{dt}=\int-\cos^2\text{u du}=-\int\Big[\frac{1+\cos2\text{u}}{2}\Big]\text{du}$
$=-\frac{\text{u}}{2}-\frac{\sin2\text{u}}{4}$
Substitute back $\text{u}=\sin^{-1}\text{t}$ and $\text{t}=\sqrt{\text{x}}$
$=-\frac{\sin^{-1}\sqrt{\text{x}}}{2}-\frac{\sin(2\sin^{-1}\sqrt{\text{x}})}{4}$
$\int\sin^{-1}\sqrt{\text{x}}\text{dx}=\text{x}\sin^{-1}\sqrt{\text{x}}-\frac{\sin^{-1}\sqrt{\text{x}}}{2}-\frac{\sin(2\sin^{-1}\sqrt{\text{x}})}{4}$
$\sin(2\sin^{-1}\sqrt{\text{x}})=2\sqrt{\text{x}}\sqrt{1-\text{x}}$
$\int\sin^{-1}\sqrt{\text{x}}\text{dx}=\text{x}\sin^{-1}\sqrt{\text{x}}-\frac{\sin^{-1}\sqrt{\text{x}}}{2}-\frac{\sqrt{\text{x}(1-\text{x})}}{2}$
View full question & answer→Question 3625 Marks
Evaluvate the following intregals:
$\int\frac{1}{1-\tan\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{1-\tan\text{x}}\text{ dx}$
$=\int\frac{1}{1-\frac{\sin\text{x}}{\cos\text{x}}}\ \text{dx}$
$=\int\frac{\cos\text{x}}{\cos\text{x}-\sin\text{x}}\ \text{dx}$
Let $\cos\text{x}=\lambda\frac{\text{d}}{\text{dx}}(\cos\text{x}-\sin\text{x})+\mu(\cos\text{x}-\sin\text{x})+\text{v}$
$=\lambda\frac{\text{d}}{\text{dx}}(-\sin\text{x}-\cos\text{x})+\mu(\cos\text{x}-\sin\text{x})+\text{v}$
$\cos\text{x}=\sin(-\lambda-\mu)+\cos\text{x}(-\lambda+\mu)+\text{v}$
Compairing the cooefficients of $\cos\text{x}\ \&\sin\text{x}$ on the both the sides,
$-\lambda-\mu=0\ ...(1)$
$-\lambda+\mu=1\ ...(2)$
$\text{v}=0\ ...(3)$
Equation (1), (2), (3) gives
$\lambda=-\frac{1}{2},\mu=\frac{1}{2},\text{v}=0$
$\text{I}=\int\frac{-\frac{1}{2}(-\sin\text{x}-\cos\text{x})+\frac{1}{2}(\cos\text{x}-\sin\text{x}_)}{(\cos\text{x}-\sin\text{x})}\ \text{dx}$
$=\frac{1}{2}\int\frac{(\cos\text{x}+\sin\text{x})}{(\sin\text{x}-\cos\text{x})}\ \text{dx}+\frac{1}{2}\int\ \text{dx}$
$\text{I}=-\frac{1}{2}\log|\cos\text{x}-\sin\text{x}|+\frac{1}{2}\text{x}+\text{C}$
$\text{I}=\frac{1}{2}\text{x}-\frac{1}{2}\log|\cos\text{x}-\sin\text{x}|+\text{C}$
View full question & answer→Question 3635 Marks
Evaluvate the following intregals
$\int\frac{2\text{x}+3}{\sqrt{\text{x}^2+4\text{x}+5}}\text{dx}$
AnswerLet $\text{I}=\int\frac{2\text{x}+3}{\sqrt{\text{x}^2+4\text{x}+5}}\text{dx}$
Let $2\text{x}+3=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+4\text{x}+5)+\mu$
$=\lambda(2\text{x}+4)+\mu$
$2\text{x}+3=(2\lambda)\text{x}+4\lambda+\mu$
Compairing the coefficient of like powers of x,
$2\lambda=2\ \Rightarrow\lambda=1$
$4\lambda+\mu=3\Rightarrow4(1)+\mu=3$
$\Rightarrow\mu=-1$
So, $\text{I}=\int\frac{(2\text{x}+4)-1}{\sqrt{\text{x}^2+4\text{x}+5}}\text{dx}$
$=\int\frac{(2\text{x}+4)}{\sqrt{\text{x}^2+4\text{x}+5}}\text{dx}-\int\frac{1}{\sqrt{\text{x}^2+2\text{x}(2)+(2)^2-(2)^2+5}}$
$=\int\frac{(2\text{x}+4)}{\sqrt{\text{x}^2+4\text{x}+5}}\text{dx}-\int\frac{1}{\sqrt{(\text{x}+2)^2+(1)^2}}\text{dx}$
$\text{I}=2\sqrt{\text{x}^2+4\text{x}+5}-\log\big|\text{x}+2+\sqrt{(\text{x}+2)^2+1}\big|+\text{C}$
$\Big[\text{since}, \int\frac{1}{\sqrt{\text{x}}}\text{dx}=2\sqrt{\text{x}}+\text{C},\int\frac{1}{\sqrt{\text{x}^2-\text{a}^2}}\text{dx}=\log\big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\big|+\text{C}\Big]$
$\text{I}=2\sqrt{\text{x}^2+4\text{x}+5}-3\log\big|\text{x}+2+\sqrt{\text{x}^2+4\text{x}+5}\big|+\text{C}$
View full question & answer→Question 3645 Marks
Evaluate the following integrals:
$\int^\limits{1}_0\big(\cos^{-1}\text{x}\big)^2\text{dx}$
AnswerWe have,
$\int^\limits{1}_0\big(\cos^{-1}\text{x}\big)^2\text{dx}=(\cos^{-1}\text{x}\big)^2\int\limits^1_0\text{dx}-\int\limits^1_0\big(\int\text{dx}\big)\frac{\text{d}\big(\cos^{-1}\text{x}\big)^2}{\text{dx}}\text{ dx}$
$=\Big[\text{x}\big(\cos^{-1}\text{x}\big)^2\Big]^1_0+\int^\limits{1}_0\frac{\text{x}\cdot2\cos^{-1}}{\sqrt{1-\text{x}^2}}\text{ dx}$
Now,
Let $\cos^{-1}\text{x}=\text{t}\Rightarrow-\frac{1}{\sqrt{1-\text{x}^2}}\text{ dx}=\text{dt}$
When $\text{x}=0\Rightarrow\text{t}=\frac{\pi}{2}$
$\text{x}=1\Rightarrow\text{t}=0$
$\therefore\ \int^\limits{1}_0\frac{2\text{ x}\cos^{-1}}{\sqrt{1-\text{x}^2}}\text{ dx}=-2\int^0_\limits{\frac{\pi}{2}}\text{t}\cos\text{t dt}=2\int^\limits{\frac{\pi}{2}}_0\text{t}\cos\text{t dt}$
$=2\Big[\text{t}\int\cos\text{t dt}-\int\big(\cos\text{t dt}\big)\frac{\text{dt}}{\text{dt}}\text{ dt}\Big]^{\frac{\pi}{2}}_0$
$=2\Big[\text{t}\sin\text{t}-\int\sin\text{t dt}\Big]^{\frac{\pi}{2}}_0$
$=2\Big[\text{t}\sin\text{t}+\cos\text{t}\Big]^{\frac{\pi}{2}}_0$
$=2\Big[\frac{\pi}{2}-1\Big]$
$\int^\limits{1}_0\big(\cos^{-1}\text{x}\big)^2\text{dx}=\Big[\text{x}\big(\cos^{-1}\text{x}\big)^2\Big]^1_0+\int^\limits{1}_0\frac{\text{x}\cdot2\cos^{-1}}{\sqrt{1-\text{x}^2}}\text{ dx}\\=\Big[\text{x}\big(\cos^{-1}\text{x}\big)^2\Big]^1_0+2\Big(\frac{\pi}{2}-1\Big)$
$=0-0+2\Big(\frac{\pi}{2}-1\Big)$
$=(\pi-2)$
View full question & answer→Question 3655 Marks
Evaluate the following intregals:
$\int\frac{1}{13+3\cos\text{x}+4\sin\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{13+3\cos\text{x}+4\sin\text{x}}\ \text{dx}$
Putting $\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}},\sin\text{x}=\frac{2\tan\Big(\frac{\text{x}}{2}\Big)}{1+\tan^2\Big(\frac{\text{x}}{2}\Big)}$
$\therefore\text{I}=\int\frac{1}{13+3\Bigg(\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)+4\times2\frac{\tan\Big(\frac{\text{x}}{2}\Big)}{1+\tan^2\Big(\frac{\text{x}}{2}\Big)}}\ \text{dx}$
$=\int\frac{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}{13\Big(1+\tan^2\frac{\text{x}}{2}\Big)+3-3\tan^2\frac{\text{x}}{2}+16+8\tan\Big(\frac{\text{x}}{2}\Big)}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{13\tan^2\frac{\text{x}}{2}-3\tan^2\frac{\text{x}}{2}+16+8\tan\Big(\frac{\text{x}}{2}\Big)}\ \text{dx}$
$==\int\frac{\sec^2\frac{\text{x}}{2}}{10\tan^2\Big(\frac{\text{x}}{2}\Big)+8\tan\Big(\frac{\text{x}}{2}\Big)+16}\ \text{dx}$
Let $\tan\Big(\frac{\text{x}}{2}\Big)=\text{t}$
$\Rightarrow\frac{1}{2}\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=\text{dt}$
$\Rightarrow\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=2\text{dt}$
$\therefore\text{I}=\int\frac{2\text{dt}}{10\text{t}^2+8\text{t}+16}$
$=\int\frac{\text{dt}}{5\text{t}^2+4\text{t}+8}$
$=\frac{1}{5}\int\frac{\text{dt}}{\text{t}^2+\frac{4}{5}\text{t}+\frac{8}{5}}$
$=\frac{1}{5}\int\frac{\text{dt}}{\text{t}^2+\frac{4}{5}\text{t}+\Big(\frac{2}{5}\Big)^2-\Big(\frac{2}{5}\Big)^2+\frac{8}{5}}$
$=\frac{1}{5}\int\frac{\text{dt}}{\Big(\text{t}+\frac{2}{5}\Big)^2-\frac{4}{25}+\frac{8}{5}}$
$=\frac{1}{5}\int\frac{\text{dt}}{\Big(\text{t}+\frac{2}{5}\Big)^2+\frac{-4+40}{25}}$
$=\frac{1}{5}\int\frac{\text{dt}}{\Big(\text{t}+\frac{2}{5}\Big)+\Big(\frac{6}{5}\Big)^2}$
$=\frac{1}{5}\times\frac{5}{6}\tan^{-1}\Bigg(\frac{\text{t}+\frac{2}{5}}{\frac{6}{5}}\Bigg)+\text{C}$
$=\frac{1}{6}\tan^{-1}\Big(\frac{5\text{t}+2}{6}\Big)+\text{C}$
$=\frac{1}{6}\tan^{-1}\bigg(\frac{5\tan\frac{\text{x}}{2}+2}{6}\bigg)+\text{C}$
View full question & answer→Question 3665 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^\text{b}_{\text{a}}\cos\text{x dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=\text{a},\text{ b}=\text{b},\text{ f(x)}=\cos\text{x},\text{ h}=\frac{\text{b}-\text{a}}{\text{n}}$
Therefore, $\text{I}=\int\limits^\text{b}_{\text{a}}\cos\text{x dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\ ....\ +\text{f}\big\{\text{a}+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\cos(\text{a})+\cos(\text{a}+\text{h})+\ ....+\ \cos\big\{\text{a}+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Bigg[\frac{\cos\big\{\text{a}+(\text{n}-1)\frac{\text{h}}{2}\big\}\sin\frac{\text{nh}}{2}}{\sin\frac{\text{h}}{2}}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg[\frac{\frac{\text{h}}{2}}{\sin\frac{\text{h}}{2}}2\cos\Big(\text{a}+\frac{\text{b}-\text{a}}{2}-\frac{\text{h}}{2}\Big)\sin\Big(\frac{\text{b}-\text{a}}{2}\Big)\Bigg]$ (Using nh = b - a)
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\text{h}}{2}}{\sin\frac{\text{h}}{2}}\times\lim\limits_{\text{h}\rightarrow0}2\cos\Big(\frac{\text{a}+\text{b}}{2}-\frac{\text{h}}{2}\Big)\sin\Big(\frac{\text{b}-\text{a}}{2}\Big)$
$=2\cos\Big(\frac{\text{a}+\text{b}}{2}\Big)\sin\Big(\frac{\text{b}-\text{a}}{2}\Big)$
$=\sin\text{b}-\sin\text{a}$ $\Big[\text{Since},2\cos\text{A}\sin\text{B}=\sin(\text{A}+\text{B})-\sin(\text{A}-\text{B})\Big]$
View full question & answer→Question 3675 Marks
Evaluate the following integrals:
$\int\text{x}^2\text{e}^{\text{x}^3}\cos\text{x}^3\text{dx}$
AnswerGiven integral is,
$\text{I}=\int\text{x}^2\text{e}^{\text{x}^3}\cos(\text{x}^3)\text{dx}$
Let $x^3 = t$
$\Rightarrow3\text{x}^2\text{dx}=\text{dt}$
$\Rightarrow\text{x}^2\text{dx}=\frac{\text{dt}}{3}$
Integral becomes,
$\frac{1}{3}\int\text{e}^\text{t}\cos\text{t dt}$
$=\frac{1}{3}\text{I}\ \dots(1)$
Where, $\text{I}=\int\text{e}^\text{t}\cos\text{t dt}$
$\text{I}=\int\text{e}^\text{t}\cos\text{t dt}$
Considering cos t as first and $e^t$ as second function
$\text{I}=\cos\text{t e}^\text{t}-\int-\sin\text{t e}^\text{t}\text{dt}$
$\Rightarrow\text{I}=\text{e}^\text{t}\cos\text{t}+\int\sin\text{t }\text{e}^\text{t}\text{dt}$
Again considering sin t as first and $e^t$ as second function
$\text{I}=\text{e}^\text{t}\cos\text{t}+\sin\text{t }\text{e}^\text{t}-\int\cos\text{t e}^\text{t}\text{dt}$
$\Rightarrow\text{I}=\text{e}^\text{t}\cos\text{t}+\sin\text{t e}^\text{t}-1$
$\Rightarrow2\text{I}=\text{e}^\text{t}(\sin\text{t}+\cos\text{t})$
$\Rightarrow\text{I}=\frac{\text{e}^\text{t}}{2}(\sin\text{t}+\cos\text{t})$
$\therefore\ \int\text{x}^2\text{e}^{\text{x}^3}\cos(\text{x}^3)\text{dx}=\frac{1}{3}\Big[\frac{\text{e}^\text{t}}{2}(\sin\text{t}+\cos\text{t})\Big]+\text{C}$ [From (1)]
$=\frac{\text{e}^{\text{x}^3}}{6}(\sin\text{x}^3+\cos\text{x}^3)+\text{C}$
View full question & answer→Question 3685 Marks
$\int\text{e}^{-3\text{x}}\cos^3\text{x dx}$
AnswerLet $\text{I}=\int\text{e}^{-3\text{x}}\cos^3\text{x dx}$
$=\int\text{e}^{-3\text{x}}\Big(\frac{\cos3\text{x}+3\cos\text{x}}{4}\Big)\text{dx}$
$=\frac{1}{4}\int\big(\text{e}^{-3\text{x}}\cos3\text{x}+\text{e}^{-3\text{x}}\cos\text{x}\big)\text{dx}$
$=\frac{1}{4}(\text{I}_1+\text{I}_2)$
$\text{I}_1=\int\text{e}^{-3\text{x}}\cos3\text{x dx}$
$=\text{e}^{-3\text{x}}\int\cos3\text{x dx}-\int\big((\text{e}^{-3\text{x}})\int\cos3\text{x dx}\big)\text{dx}$
$=\text{e}^{-3\text{x}}\frac{\sin3\text{x}}{3}-\int-3\text{e}^{-3\text{x}}\frac{\sin3\text{x}}{3}\text{dx}$
$=\text{e}^{-3\text{x}}\frac{\sin3\text{x}}{3}+\int\text{e}^{-3\text{x}}\sin3\text{x dx}$
$=\text{e}^{-3\text{x}}\frac{\sin3\text{x}}{3}+\text{e}^{-3\text{x}}\int\sin3\text{x dx}-\int\big((\text{e}^{-3\text{x}})\int\sin3\text{x dx})\text{dx}$
$=\text{e}^{-3\text{x}}\frac{\sin3\text{x}}{3}-\text{e}^{-3\text{x}}\frac{\cos3\text{x}}{3}-\int\text{e}^{-3\text{x}}\cos3\text{x dx}$
$=\text{e}^{-3\text{x}}\frac{3\sin}{3}-\text{e}^{-3\text{x}}\frac{\cos3\text{x}}{3}-\text{I}_1$
$\Rightarrow\ 2\text{I}_1=\frac{\text{e}^{-3\text{x}}}{3}(\sin3\text{x}-\cos3\text{x})$
$\Rightarrow\ \text{I}_1=\frac{\text{e}^{-3\text{x}}}{6}(\sin3\text{x}-\cos3\text{x})+\text{C}_1$
Similarly $\text{I}_2=\int\text{e}^{-3\text{x}}\cos\text{x dx}=\frac{\text{e}^{-3\text{x}}}{10}(\sin3\text{x}-3\cos3\text{x})+\text{C}_2$
$\Rightarrow\ \text{I}=\frac{1}{4}\bigg[\frac{\text{e}^{-3\text{x}}}{6}(\sin3\text{x}-\cos3\text{x})+\frac{\text{e}^{-3\text{x}}}{10}(\sin3\text{x}-3\cos3\text{x})\bigg]+\text{C}$
View full question & answer→Question 3695 Marks
Evaluate the following intregals:
$\int\frac{1}{5-4\cos\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{5-4\cos\text{x}}\ \text{dx}$
Putting $\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$=\text{I}=\int\frac{1}{5-4\Bigg(\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)}\ \text{dx}$
$=\int\frac{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}{5\Big(1+\tan^2\frac{\text{x}}{2}\Big)-4+4\tan^2\frac{\text{x}}{2}}$
$=\int\frac{\sec^2\Big(\frac{\text{x}}{2}\Big)}{9\tan^2\frac{\text{x}}{2}+1}\ \text{dx}$
Let $\tan\big(\frac{\text{x}}{2}\big)=\text{t}$
$\Rightarrow\frac{1}{2}\sec^2\big(\frac{\text{x}}{2}\big)\text{dx}=\text{dt}$
$\Rightarrow\sec^2\big(\frac{\text{x}}{2}\big)\text{dx}=2\text{dt}$
$\therefore\text{I}=2\int\frac{\text{dt}}{9\text{t}^2+1}$
$=\frac{2}{9}\int\frac{\text{dt}}{\text{t}^2+\frac{1}{9}}$
$=\frac{2}{9}\int\frac{\text{dt}}{\text{t}^2+\big(\frac{1}{3}\big)^2}$
$=\frac{2}{9}\times3\tan^{-1}\bigg(\frac{\text{t}}{\frac{1}{3}}\bigg)+\text{C}$
$=\frac{2}{3}\tan^{-1}(3\text{t})+\text{C}$
$=\frac{2}{3}\tan^{-1}\big(3\tan\frac{\text{x}}{2}\big)+\text{C}$
View full question & answer→Question 3705 Marks
Evaluvate the following intregals:
$\int\frac{5\text{x}+3}{\sqrt{\text{x}^2+4\text{x}+10}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{5\text{x}+3}{\sqrt{\text{x}^2+4\text{x}+10}}\ \text{dx}$
Consider,
$5\text{x}+3=\text{A}\frac{\text{d}}{\text{dx}}(\text{x}^2+4\text{x}+10)+\text{B}$
$\Rightarrow5\text{x}+3=\text{A}(2\text{x}+4)+\text{B}$
$\Rightarrow5\text{x}+3=(2\text{A})\text{x}+4\text{A}+\text{B}$
Equating coefficient of like terms
$2\text{A}=5$
$\Rightarrow\text{A}=\frac{5}{2}$
And
$4\text{A}+\text{B}=3$
$\Rightarrow4\times\frac{5}{2}+\text{B}=3$
$\Rightarrow\text{B}=-7$
$\therefore\text{I}=\frac{5}{2}\int\frac{(2\text{x}+4)\text{dx}}{\sqrt{\text{x}^2+4\text{x}+10}}-7\int\frac{\text{dx}}{\sqrt{\text{x}^2+4\text{x}+10}}$
$=\frac{5}{2}\int\frac{(2\text{x}+4)\text{dx}}{\sqrt{\text{x}^2+4\text{x}+10}}-7\int\frac{\text{dx}}{\sqrt{\text{x}^2+4\text{x}+4-4+10}}$
$=\frac{5}{2}\int\frac{(2\text{x}+4)\text{dx}}{\sqrt{\text{x}^2+4\text{x}+10}}-7\int\frac{\text{dx}}{\sqrt{(\text{x}+2})^2+(\sqrt{6})^2}$
Putting, $\text{x}^2+4\text{x}+10=\text{t}$
$\Rightarrow(2\text{x}+4)\text{dx}=\text{dt}$
$\text{I}=\frac{5}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}-7\log\big|(\text{x}+2)^2+6\big|+\text{C}$
$=\frac{5}{2}\int\text{t}^{-\frac{1}{2}}\text{dt}-7\log\big|\text{x}+2+\sqrt{\text{x}^2+4\text{x}+10}\big|+\text{C}$
$=\frac{5}{2}\times2\sqrt{\text{t}}-7\log\big|\text{x}+2+\sqrt{\text{x}^2+4\text{x}+10}\big|+\text{C}$
$=5\sqrt{\text{x}^2+4\text{x}+10}-7\log\big|\text{x}+2+\sqrt{\text{x}^2+4\text{x}+10}\big|+\text{C}$
View full question & answer→Question 3715 Marks
Evaluate the following integrals:$\int\frac{\text{x}^2}{\text{x}^2+7\text{x}+10}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{\text{x}^2+7\text{x}+10}\text{ dx}$ $=\int\Big\{1-\frac{7\text{x}+10}{\text{x}^2+7\text{x}+10}\Big\}\text{dx}$ $=\text{x}-\int\frac{7\text{x}+10}{\text{x}^2+7\text{x}+10}\text{ dx}+\text{C}_1\ ....(1)$ $\text{I}_1=\int\frac{7\text{x}+10}{\text{x}^2+7\text{x}+10}\text{ dx}$ Let $7\text{x}+10=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2+7\text{x}+10\big)+\mu$ $=\lambda(2\text{x}+7)+\mu$ $7\text{x}+10=(2\lambda)\text{x}+7\lambda+\mu$Comparing the coefficients of like powers of x,
$7=2\lambda\Rightarrow\lambda=\frac{7}{2}$ $7\lambda+\mu=10\Rightarrow7\Big(\frac{7}{2}\Big)+\mu=10$ $\mu=-\frac{29}{2}$ So, $\text{I}_1=\int\frac{\frac{7}{2}(2\text{x}+7)-\frac{29}{2}}{\text{x}^2+7\text{x}+10}\text{ dx}$ $\text{I}_1=\frac{7}{2}\int\frac{(2\text{x}+7)}{\text{x}^2+7\text{x}+10}\text{ dx}-\frac{29}{2}\int\frac{1}{\text{x}^2-2\text{x}\big(\frac{7}{2}\big)+\big(\frac{7}{2}\big)^2-\big(\frac{7}{2}\big)^2+10}\text{ dx}$ $\text{I}_1=\frac{7}{2}\int\frac{2\text{x}+7}{\text{x}^2+7\text{x}+10}\text{ dx}-\frac{29}{2}\int\frac{1}{\big(\text{x}+\frac{7}{2}\big)^2-\big(\frac{3}{2}\big)^2}\text{ dx}$ $\text{I}_1=\frac{2}{7}\log\big|\text{x}^2+7\text{x}+10\big|-\frac{29}{2}\times\frac{1}{2\big(\frac{3}{2}\big)}\log\bigg|\frac{\text{x}+\frac{7}{2}-\frac{3}{2}}{\text{x}+\frac{7}{2}+\frac{3}{2}}\bigg|+\text{C}_2$ $\Big[\text{since},\int\frac{1}{\text{x}^2-\text{a}^2}\text{ dx}=\frac{1}{2\text{a}}\log\Big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big|+\text{C}\Big]$ $\text{I}_1=\frac{7}{2}\log\big|\text{x}^2+7\text{x}+10\big|-\frac{29}{6}\log\Big|\frac{\text{x}+2}{\text{x}+5}\Big|+\text{C}_2\ ....(2)$ Using equation (1) and (2) $\text{I}=\text{x}-\frac{7}{2}\log\big|\text{x}^2+7\text{x}+10\big|+\frac{29}{6}\log\Big|\frac{\text{x}+2}{\text{x}+5}\Big|+\text{C}$
View full question & answer→Question 3725 Marks
Evaluate the follwing intregals:
$\int\frac{1}{\text{x}(\text{x}^4-1)}\ \text{dx}$
AnswerWe have,
$\text{I}=\int\frac{\text{dx}}{\text{x}(\text{x}^4-1)}$
$=\int\frac{\text{x}^3\text{dx}}{\text{x}^4(\text{x}^4-1)}$
Putting $\text{x}^4=\text{t}$
$\Rightarrow4\text{x}^3\text{dx}=\text{dt}$
$\Rightarrow\text{x}^3\text{dx}=\frac{\text{dt}}{4}$
$\therefore\text{I}=\frac{1}{4}\int\frac{\text{dt}}{\text{t}(\text{t}-1)}$
Let $\frac{1}{\text{t}(\text{t}-1)}=\frac{\text{A}}{\text{t}}+\frac{\text{B}}{\text{t}-1}$
$\rightarrow\frac{1}{\text{t}(\text{t}-1)}=\frac{\text{A}(\text{t}-1)+\text{B}\text{t}}{(\text{t}-1)}$
$\Rightarrow1=\text{A}(\text{t}-1)+\text{Bt}$
Putting t - 1 = 0
⇒ t = 1
$\therefore$ 1 = A × 0 + B (1)
⇒ B = 1
Putting t = 0
$\therefore$ 1 = A (0 - 1) + B × 0
⇒ A = -1
$\therefore$ $\text{I}=-\frac{1}{4}\int\frac{\text{dt}}{\text{t}}+\frac{1}{4}\int\frac{\text{dt}}{\text{t}-1}$
$=-\frac{1}{4}\log|\text{t}|+\frac{1}{4}\log|\text{t}-1|+\text{C}$
$=\frac{1}{4}\log\Big|\frac{\text{t}-1}{\text{t}}\Big|+\text{C}$
$=\frac{1}{4}\log\Big|\frac{\text{x}^2-1}{\text{x}^4}\Big|+\text{C}$
View full question & answer→Question 3735 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\tan\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\tan\text{x}}}\text{ dx}\ ....(\text{i})$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\tan\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\cot\text{x}}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\Big(\frac{1}{1+\sqrt{\tan\text{x}}}+\frac{1}{1+\sqrt{\cot\text{x}}}\Big)\text{dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\bigg(\frac{1+\sqrt{\cot\text{x}}+1+\sqrt{\tan\text{x}}}{1+\sqrt{\cot\text{x}}+1+\sqrt{\tan\text{x}}+\sqrt{\tan\text{x}\cot\text{x}}}\bigg)\text{dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\text{dx}=\Big[\text{x}\Big]^{\frac{\pi}{3}}_\frac{\pi}{6}$
$=\frac{\pi}{3}-\frac{\pi}{6}$
$\therefore\ 2\text{I}=\frac{\pi}{6}$
Hence, $\text{I}=\frac{\pi}{12}$
View full question & answer→Question 3745 Marks
Evaluate the following:
$\int\frac{\text{x}^{\frac{1}{2}}}{1+\text{x}^{\frac{3}{4}}}\text{dx}$
Hint: Put $x = z^4$
AnswerLet $\text{I}=\int\frac{\text{x}^{\frac{1}{2}}}{1+\text{x}^{\frac{3}{4}}}\text{dx}$
Put $\text{x}=\text{t}^4\Rightarrow\text{dx}=4\text{t}^3\text{dt}$
$\therefore\ \text{I}=4\int\frac{\text{t}^2(\text{t}^3)}{1+\text{t}^3}\text{dt}=4$ $\Big(\text{t}^2-\frac{\text{t}^2}{1+\text{t}^3}\Big)\text{dt}$
$\text{I}=4\int\text{t}^2\text{dt}-4\int\frac{\text{t}^2}{1+\text{t}^3}$
$\text{I}=\text{I}_1+\text{I}_2$
$\text{I}_1=4\int\text{t}^2\text{dt}=4\cdot\frac{\text{t}^3}{3}+\text{C}_1=\frac{4}{3}\text{x}^{\frac{3}{4}}+\text{C}_1$
Now, $\text{I}_2=4\int\frac{\text{t}^2}{1+\text{t}^3}\text{dt}$
Again, Put $1+\text{t}^3=\text{z}\Rightarrow3\text{t}^2\text{dt}=\text{dz}$
$\Rightarrow\ \text{t}^2\text{dt}=\frac{1}{3}\text{dz}=\frac{4}{3}\int\frac{1}{\text{z}}\text{dz}$
$=\frac{4}{3}\log|\text{z}|+\text{C}_2=\frac{4}{3}\log\big|(1+\text{t}^3)\big|+\text{C}_2$
$=\frac{4}{3}\log\Big|(1+\text{x}^{\frac{3}{4}})\Big|+\text{C}_2$
$\therefore\ \text{I}=\frac{4}{3}\text{x}^{\frac{3}{4}}+\text{C}_1-\frac{4}{3}\log\Big|(1+\text{x}^{\frac{3}{4}})\Big|-\text{C}_2$
$=\frac{4}{3}\Big(\text{x}^{\frac{3}{4}}-\log\Big|(1+\text{x}^{\frac{3}{4}})\Big|\Big)+\text{C}$ $[\because\text{C}=\text{C}_1-\text{C}_2]$
View full question & answer→Question 3755 Marks
Evaluate the following integrals:$\int\frac{2\text{x}-3}{\text{x}^2+6\text{x}+13}\text{ dx}$
AnswerLet $\text{I}=\int\frac{2\text{x}-3}{\text{x}^2+6\text{x}+13}\text{ dx}$
Let $2\text{x}-3=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2+6\text{x}+13\big)+\mu$
$=\lambda(2\text{x}+6)+\mu$
$2\text{x}-3=(2\lambda)\text{x}+(6\lambda+\mu)$
Comparing the coefficients of like powers of x,
$2\lambda=2\Rightarrow\lambda=1$
$6\lambda+\mu=-3\Rightarrow6(1)+\mu=-3$
$\mu=-9$
So, $\text{I}=\int\frac{1(2\text{x}+6)-9}{\text{x}^2+6\text{x}+13}\text{ dx}$
$\text{I}=\int\frac{2\text{x}+6}{\text{x}^2+6\text{x}+13}\text{ dx}-9\int\frac{1}{\text{x}^2+2\text{x}(3)+(3)^2-(3)^2+13}\text{ dx}$
$\text{I}=\int\frac{2\text{x}+6}{\text{x}^2+6\text{x}+13}\text{ dx}-9\int\frac{1}{(\text{x}+3)^2+(2)^2}\text{ dx}$
$\text{I}=\log\big|\text{x}^2+6\text{x}+13\big|-9\times\frac{1}{2}\tan^{-1}\Big(\frac{\text{x}+3}{2}\Big)+\text{C}$ $\Big[\text{Since }\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
$\text{I}=\log\big|\text{x}^2+6\text{x}+13\big|-\frac{9}{2}\tan^{-1}\Big(\frac{\text{x}+3}{2}\Big)+\text{C}$
View full question & answer→Question 3765 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^2_{1}\text{x}^2\text{ dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=1,\text{ b}=2,\text{ f(x)}=\text{x}^2,\text{ h}=\frac{2-1}{\text{n}}=\frac{1}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_{1}\text{x}^2\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(1)+\text{f}(1+\text{h})\ ....\ +\text{f}\big(1+(\text{n}-1)\text{h}\big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(1)+(\text{h}+1)^2+\ ....\ +\big((\text{n}-1)\text{h}+1\big)^2\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}^2\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2\big\}\\+2\text{h}\big\{1+2+3+\ ....\ +(\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{n}+\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}+2\text{h}\frac{\text{n}(\text{n}-1)}{2}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\bigg[\text{n}+\frac{(\text{n}-1)(2\text{n}-1)}{6\text{n}}+\text{n}-1\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\bigg\{2+\frac{1}{6}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{1}{\text{n}}\bigg\}$
$=2+\frac{1}{3}$
$=\frac{7}{3}$
View full question & answer→Question 3775 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\big(2\log\cos\text{x}-\log\sin2\text{x}\big)\text{dx}$
AnswerWe have,
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\big(2\log\cos\text{x}-\log\sin2\text{x}\big)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\big(\log\cos^2\text{x}-\log\sin2\text{x}\big)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log\frac{\cos^2\text{x}}{\sin\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log\frac{\cos^2\text{x}}{2\sin\text{x}\cdot\cos\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log\frac{\cos\text{x}}{2\sin\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\big(\log\cos\text{x}-\log\sin\text{x}-\log2\big)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log\cos\text{x dx}-\int\limits^{\frac{\pi}{2}}_0\log\sin\text{x dx}-\int\limits^{\frac{\pi}{2}}_0\log2$
We know that
$\int\limits^{\frac{\pi}{2}}_0\log\cos\text{x dx}=\int\limits^{\frac{\pi}{2}}_0\log\sin\text{x dx}\ ...(\text{i})$
Hence from equation (i)
$\text{I}=-\int\limits^{\frac{\pi}{2}}_0\log2=-\frac{\pi}{2}\log2$
View full question & answer→Question 3785 Marks
Evaluate the following definite integrals:
$\int_{1}^\limits{4}\frac{\text{x}^2+\text{x}}{\sqrt{2\text{x}+1}}\text{ dx}$
AnswerLet $\text{I}=\int_{1}^\limits{4}\frac{\text{x}^2+\text{x}}{\sqrt{2\text{x}+1}}\text{ dx}$
Let $2\text{x}+1=\text{u}$
$\Rightarrow\text{x}=\frac{\text{u}-1}{2}$
$\Rightarrow\text{dx}=\frac{\text{du}}{2}$
$\therefore\ \text{I}=\int\frac{\big(\frac{\text{u}-1}{2}\big)^2+\frac{\text{u}-1}{2}}{\sqrt{\text{u}}}\frac{\text{du}}{2}$
$\Rightarrow\text{I}=\frac{1}{8}\int\frac{\text{u}^2+1-2\text{u}+2\text{u}-2}{\sqrt{\text{u}}}\text{ du}$
$\Rightarrow\text{I}=\frac{1}{8}\int\frac{(\text{u}^2-1)}{\sqrt{\text{u}}}\text{ du}$
$\Rightarrow\text{I}=\frac{1}{8}\int\Big(\text{u}^{\frac{3}{2}}-\text{u}^{-\frac{1}{2}}\Big)\text{du}$
$\Rightarrow\text{I}=\frac{1}{8}\bigg[\frac{2\text{u}^{\frac{5}{2}}}{5}-\frac{2\text{u}^{\frac{1}{2}}}{1}\bigg]$
$\Rightarrow\text{I}=\frac{1}{8}\Big[\frac{2}{5}\times243-6-\frac{2}{5}\times9\sqrt{3}+2\sqrt{3}\Big]$
$\Rightarrow\text{I}=\frac{1}{8}\Big[\frac{456}{5}-\frac{8\sqrt{3}}{5}\Big]$
$\Rightarrow\text{I}=\frac{57-\sqrt{3}}{5}$
View full question & answer→Question 3795 Marks
Evaluate the following integrals:$\int\frac{\text{ax}^3+\text{bx}}{\text{x}^4+\text{c}^2}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{ax}^3+\text{bx}}{\text{x}^4+\text{c}^2}\text{ dx}$ Let $\text{ax}^3+\text{bx}=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^4+\text{c}^2\big)+\mu$ $\text{ax}^3+\text{bx}=\lambda\big(4\text{x}^3\big)+\mu$ Comparing the coefficients of like powers of x, $4\lambda=\text{a}\Rightarrow\lambda=\frac{\text{a}}{4}$ $\mu=0\Rightarrow\mu=0$ So, $\text{I}=\int\frac{\frac{\text{a}}{4}(4\text{x}^3)+\text{bx}}{\text{x}^4+\text{c}^2}\text{ dx}$ $\text{I}=\frac{\text{a}}{4}\int\frac{4\text{x}^3}{\text{x}^4+\text{c}^2}\text{ dx}+\text{b}\int\frac{\text{x}}{(\text{x}^2)^2+\text{c}^2}\text{ dx}$ $\text{I}=\frac{\text{a}}{4}\int\frac{4\text{x}^3}{\text{x}^4+\text{c}^2}\text{ dx}+\frac{\text{b}}{2}\int\frac{2\text{x}}{(\text{x}^2)^2+\text{c}^2}\text{ dx}$ $\text{I}=\frac{\text{a}}{4}\log\big|\text{x}^4+\text{c}^2\big|+\frac{\text{b}}{2}\text{I}_1\ ....(1)$Now,
$\text{I}_1=\int\frac{2\text{x}}{(\text{x}^2)^2+\text{c}^2}\text{ dx}$ Put $\text{x}^2=\text{t}$ $\Rightarrow2\text{x}\text{ dx}=\text{dt}$ $\text{I}_1=\int\frac{1}{(\text{t})^2+\text{c}^2}\text{ dx}$ $\text{I}_1=\frac{1}{\text{c}}\tan^{-1}\Big(\frac{\text{t}}{\text{c}}\Big)+\text{C}_1\ ....(2)$ Using equation (2) in equation (1), $\text{I}=\frac{\text{a}}{4}\log\big|\text{x}^4+\text{c}^4\big|+\frac{\text{b}}{2\text{c}}\tan^{-1}\Big(\frac{\text{x}^2}{\text{c}}\Big)+\text{K}$ K = Integration constant.
View full question & answer→Question 3805 Marks
Evaluate the following integrals:
$\int\frac{\log\big(1+\frac{1}{\text{x}}\big)}{\text{x}(1+\text{x})}\text{dx}$
AnswerLet I $=\int\frac{\log\big(1+\frac{1}{\text{x}}\big)}{\text{x}(1+\text{x})}\text{dx}\ .....(1)$
Let $\log\Big(1+\frac{1}{\text{x}}\Big)=\text{t}$ then,
$\text{d}\Big[\log\Big(1+\frac{1}{\text{x}}\Big)\Big]=\text{dt}$
$\Rightarrow\frac{1}{1+\frac{1}{\text{x}}}\times\frac{-1}{\text{x}^2}\text{dx}=\text{dt}$
$\Rightarrow\frac{1}{\frac{\text{x}+1}{\text{x}}}\times\frac{-1}{\text{x}^2}\text{dx}=\text{dt}$
$\Rightarrow\frac{-\text{x}}{\text{x}^2(\text{x}+1)}\text{dx}=-\text{dt}$
$\Rightarrow\frac{\text{dx}}{\text{x}(\text{x}+1)}=-\text{dt}$
Putting $\log\Big(1+\frac{1}{\text{x}}\Big)=\text{t}$ and $\frac{\text{dx}}{\text{x}(\text{x}+1)}=-\text{dt}$ in equation (1), we get
$\text{I}=\int\text{tx}-\text{dt}$
$=-\frac{\text{t}^2}{2}+\text{C}$
$=-\frac{1}{2}\Big[\log\Big(1+\frac{1}{\text{x}}\Big)\Big]^2+\text{C}$
$\therefore\text{I}=-\frac{1}{2}\Big[\log\Big(1+\frac{1}{\text{x}}\Big)\Big]^2+\text{C}$
View full question & answer→Question 3815 Marks
Evaluate the following integrals:
$\int\frac{\text{dx}}{(\text{x}^2+1)(\text{x}^2+4)}$
AnswerLet $\frac{1}{(\text{x}^2+1)(\text{x}^2+4)}=\frac{\text{Ax}+\text{B}}{(\text{x}^2+1)}+\frac{\text{Cx}+\text{D}}{\text{x}^2+4}$
$\Rightarrow1=(\text{Ax}+\text{B})(\text{x}^2+4)+(\text{Cx}+\text{D})(\text{x}^2+1)$
$=(\text{A}+\text{C})\text{x}^3+(\text{B}+\text{D})\text{x}^2+(4\text{A}+\text{C})\text{x}+4\text{B}+\text{D}$
Equating similar terms, we get,
A + C = 0, B + D = 0, 4A + C = 0, 4B + D = 1
Solving, we get, $\text{A}=0,\text{B}=\frac{1}{3},\text{C}=0,\text{D}=-\frac{1}{3}$
Thus,
$\text{I}=\int\frac{\frac{1}{3}\text{dx}}{(\text{x}^2+1)}-\int\frac{\frac{1}{3}\text{dx}}{(\text{x}^2+4)}$
$=\frac{1}{3}\tan^{-1}\text{x}-\frac{1}{6}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$ $\big[\because\int\frac{\text{dx}}{\text{x}^2+\text{a}^2}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{\text{a}}\big]$
$\therefore\text{I}=\frac{1}{3}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$
View full question & answer→Question 3825 Marks
Evaluate the following integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\frac{\text{x}+\sin\text{x}}{2\cos^2\frac{\text{x}}{2}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\bigg[\frac{\text{x}}{2\cos^2\frac{\text{x}}{2}}+\frac{\sin\text{x}}{2\cos^2\frac{\text{x}}{2}}\bigg]\text{dx}$
$=\frac{1}{2}\int_{0}^\limits{\frac{\pi}{2}}\text{x}\sec^2\frac{\text{x}}{2}\text{ dx}+\int_{0}^\limits{\frac{\pi}{2}}\frac{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\text{ dx}$
$=\frac{1}{2}\bigg[\text{x}\frac{\tan\frac{\text{x}}{2}}{\frac{1}{2}}\bigg]^{\frac{\pi}{2}}_0-\frac{1}{2}\int_{0}^\limits{\frac{\pi}{2}}\frac{\tan\frac{\text{x}}{2}}{\frac{1}{2}}\text{ dx}+\int_{0}^\limits{\frac{\pi}{2}}\tan\frac{\text{x}}{2}\text{ dx}$
$=\Big[\text{x}\tan\frac{\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\int_{0}^\limits{\frac{\pi}{2}}\tan\frac{\text{x}}{2}\text{ dx}+\int_{0}^\limits{\frac{\pi}{2}}\tan\frac{\text{x}}{2}\text{ dx}$
$=\Big[\frac{\pi}{2}\tan\frac{\pi}{4}\Big]$
$=\frac{\pi}{2}\times1$
$=\frac{\pi}{2}$
View full question & answer→Question 3835 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{4}}\sec\text{x}\text{ dx}$
AnswerLet $\int_{0}^\limits{\frac{\pi}{4}}\sec\text{x}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\sec\text{x}\frac{\sec\text{x}+\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sec^2\text{x}+\sec\text{x}\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{ dx}$
Put $\text{u}=\sec\text{x}+\tan\text{x}$
$\Rightarrow\text{du}=\sec^2\text{x}+\sec\text{x}\tan\text{x dx}$
$\therefore\ \int_{0}^\limits{\frac{\pi}{4}}\frac{\sec^2\text{x}+\sec\text{x}\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{ dx}=\int\frac{\text{du}}{\text{u}}$
$\Rightarrow\text{I}=\big[\log\text{u}\big]$
$\Rightarrow\text{I}=\big[\log(\sec\text{x}+\tan\text{x})\big]^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\log\Big(\sec\frac{\pi}{4}+\tan\frac{\pi}{4}\Big)-\log(\sec0+\tan0)$
$\Rightarrow\text{I}=\log\big(\sqrt{2}+1\big)-\log1$
$\Rightarrow\text{I}=\log\big(\sqrt{2}+1\big)$
View full question & answer→Question 3845 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^{\frac{\pi}{2}}_{0}\sin\text{x dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=\frac{\pi}{2},\text{ f(x)}=\sin\text{x},\text{ h}=\frac{\frac{\pi}{2}-0}{\text{n}}=\frac{\pi}{2\text{n}}$
Therefore, $\text{I}=\int\limits^{\frac{\pi}{2}}_{0}\sin\text{x dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big(0+(\text{n}-1)\text{h}\big)\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\sin0+\sin\text{h}+\sin2\text{h}+\ ....+\ \sin(\text{n}-1)\text{h}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Bigg[\frac{\sin\big((\text{n}-1)\frac{\text{h}}{2}\big)\sin\frac{\text{nh}}{2}}{\sin\frac{\text{h}}{2}}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg[\frac{\frac{\text{h}}{2}}{\sin\frac{\text{h}}{2}}\times2\sin\Big(\frac{\pi}{2}-\frac{\text{h}}{2}\Big)\sin\frac{\pi}{4}\Bigg]$ (Using nh = b - a)
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\text{h}}{2}}{\sin\frac{\text{h}}{2}}\times\lim\limits_{\text{h}\rightarrow0}2\sin\Big(\frac{\pi}{2}-\frac{\text{h}}{2}\Big)\sin\frac{\pi}{4}$
$=2\sin\frac{\pi}{4}\sin\frac{\pi}{4}$
$=2\times\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}$
$=1$
View full question & answer→Question 3855 Marks
Evaluate the following integrals:
$\int\limits^\pi_0\Big(\frac{\text{x}}{1+\sin^2\text{x}}+\cos^7\text{x}\Big)\text{dx}$
AnswerLet $\text{I}=\int\limits^\pi_0\Big(\frac{\text{x}}{1+\sin^2\text{x}}+\cos^7\text{x}\Big)\text{dx}\ ....(\text{i})$
Then,
$\text{I}=\int\limits^\pi_0\Big(\frac{\pi-\text{x}}{1+\sin^2(\pi-\text{x})}+\cos^7(\pi-\text{x})\Big)\text{dx}$
$=\int\limits^\pi_0\Big(\frac{\pi-\text{x}}{1+\sin^2\text{x}}+\cos^7\text{x}\Big)\text{dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^\pi_0\Big(\frac{\text{x}}{1+\sin^2\text{x}}+\cos^7\text{x}+\frac{\pi-\text{x}}{1+\sin^2\text{x}}-\cos^7\text{x}\Big)\text{dx}$
$\Rightarrow2\text{I}=\pi\int\limits^\pi_0\frac{1}{1+\sin^2\text{x}}\text{ dx}$
Dividing the numerator and denominator by $\cos^2\text{x},$ we get
$2\text{I}=\pi\int\limits^\pi_0\frac{\sec^2\text{x}}{\sec^2\text{x}+\tan^2\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=\pi\int\limits^\pi_0\frac{\sec^2\text{x}}{1+2\tan^2\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=2\pi\int\limits^\pi_0\frac{\sec^2\text{x}}{1+2\tan^2\text{x}}\text{ dx}$ $\begin{bmatrix}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx},&\text{if }\text{f}(2\text{a}-\text{x})=\text{f(x)}\\0,&\text{if }\text{f}(2\text{a}-\text{x})=\text{f(x)}\end{cases}\end{bmatrix}$
Put $\tan\text{x}=\text{z}$
Then $\sec^2\text{x dx}=\text{dz}$
When $\text{x}\rightarrow0,\text{ z}\rightarrow0$
When $\text{x}\rightarrow\frac{\pi}{2},\text{ z}\rightarrow\infty$
$\therefore\ 2\text{I}=2\pi\int\limits^{\infty}_0\frac{\text{dz}}{1+\big(\sqrt{2}\text{z}\big)^2}$
$\Rightarrow2\text{I}=2\pi\times\bigg[\frac{\tan^{-1}\sqrt{2}\text{z}}{\sqrt{2}}\bigg]^{\infty}_0$
$\Rightarrow\text{I}=\frac{\pi}{\sqrt{2}}\Big(\tan^{-1}\infty-\tan^{-1}0\Big)$
$\Rightarrow\text{I}=\frac{\pi}{\sqrt{2}}\times\Big(\frac{\pi}{2}-0\Big)$
$\Rightarrow\text{I}=\frac{\pi^2}{2\sqrt{2}}$
View full question & answer→Question 3865 Marks
Evaluate the following integrals:
$\int\limits^1_{-1}|\text{x}\cos\pi\text{x}|\text{dx}$
AnswerLet $\text{I}=\int\limits^1_{-1}|\text{x}\cos\pi\text{x}|\text{dx}$
Consider $\text{f(x)}=|\text{x}\cos\pi\text{x}|$
$\text{f}(-\text{x})=\big|(\text{x})\cos\pi(-\text{x})\big|=|-\text{x}\cos\pi\text{x}|=|\text{x}\cos\pi\text{x}|\text{f(x)}$
$\therefore\ \text{I}=\int\limits^1_{-1}|\text{x}\cos\pi\text{x}|\text{dx}$
$=2\int\limits^1_{-1}|\text{x}\cos\pi\text{x}|\text{dx}$ $\begin{bmatrix}\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\begin{cases}2\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx},&\text{ if }\text{f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{f}(-\text{x})=-\text{f}(\text{x})\end{cases}\end{bmatrix}$
Now,
$|\text{x}\cos\pi\text{x}|=\begin{cases}\text{x}\cos\pi\text{x},&\text{if }0\leq\text{x}\leq\frac{1}{2}\\-\text{x}\cos\pi\text{x},&\text{if }\frac{1}{2}\text{x}\leq1\end{cases}$
$\therefore\ \text{I}=2\Bigg[\int\limits^{{1}/{2}}_0\text{x}\cos\pi\text{x dx}+\int\limits^1_{1/2}(-\text{x}\cos\text{dx})\text{dx}\Bigg]$
$=2\Big(\frac{1}{2\pi}\sin\frac{\pi}{2}-0\Big)-\frac{2}{\pi}\times\Big[\Big(-\frac{\cos\pi\text{x}}{\pi}\Big)\Big]^{\frac{1}{2}}_0\\-2\Big(\frac{1}{\pi}\sin\pi-\frac{1}{2\pi}\sin\frac{\pi}{2}\Big)+\frac{2}{\pi}\times\Big(-\frac{\cos\pi\text{x}}{\pi}\Big)\Big]^1_{\frac{1}{2}}$
$=\frac{1}{\pi}+\frac{2}{\pi^2}\Big(\cos\frac{\pi}{2}-\cos0\Big)+\frac{1}{\pi}-\frac{2}{\pi}\Big(\cos\pi-\cos\frac{\pi}{2}\Big)$
$=\frac{1}{\pi}-\frac{2}{\pi^2}+\frac{1}{\pi}+\frac{2}{\pi^2}$
$=\frac{2}{\pi}$
View full question & answer→Question 3875 Marks
Evaluate the following integrals:
$\int\frac{\text{x}}{3\text{x}^4-18\text{x}^2+11}\text{dx}$
Answer$\int\frac{\text{x dx}}{3\text{x}^4-18\text{x}^2+11}$
Let $\text{x}^2=\text{t}$
$\Rightarrow2\text{x}\text{ dx = dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
Now, $\int\frac{\text{x dx}}{3\text{x}^4-18\text{x}^2+11}$
$=\frac{1}2{}\int\frac{\text{dt}}{3\text{t}^2-18\text{t}+11}$
$=\frac{1}{3\times2}\int\frac{\text{dt}}{\text{t}^2-6\text{t}+\frac{11}{3}}$
$=\frac{1}{6}\int\frac{\text{dt}}{\text{t}^2-6\text{t}+9-9+\frac{11}{3}}$
$=\frac{1}{6}\int\frac{\text{dt}}{(\text{t}-3)^2-\frac{16}{3}}$
$=\frac{1}{6}\int\frac{\text{dt}}{(\text{t}-3)^2-\Big(\frac{4}{\sqrt{3}}\Big)^2}$
$=\frac{1}{6}\times\frac{1}{2\times\frac{4}{\sqrt{3}}}\log\Bigg|\frac{\text{t}-3-\frac{4}{\sqrt{3}}}{\text{t}-3+\frac{4}{\sqrt{3}}}\Bigg|+\text{C}$
$=\frac{\sqrt{3}}{48}\log\Bigg|\frac{\text{x}^2-3-\frac{4}{\sqrt{3}}}{\text{x}^2-3+\frac{4}{\sqrt{3}}}\Bigg|+\text{C}$
View full question & answer→Question 3885 Marks
Evaluvate the following intregals:
$\int\frac{8\cot\text{x}+1}{3\cot\text{x}+2}\ \text{dx}$
AnswerLet $\text{I}=\int\Big(\frac{8\cot\text{x}+1}{3\cot\text{x}+2}\Big)\ \text{dx}$$=\int\bigg(\frac{8\frac{\cos\text{x}}{\sin\text{x}}+1}{\frac{3\cos\text{x}}{\sin\text{x}}+2}\bigg)\text{dx}$
$=\int\Big(\frac{8\cos\text{x}+\sin\text{x}}{3\cos\text{x}+2\sin\text{x}}\Big)\text{dx}$
Now, Let $8\cos\text{x}+\sin\text{x}=\text{A}(3\cos\text{x}+2\sin\text{x})+\text{B}(-3\sin\text{x}+2\cos\text{x})\ \dots(1)$
$\Rightarrow8\cos\text{x}+\sin\text{x}+\sin\text{x}=3\text{A}\cos\text{x}+2\text{A}\sin\text{x}-3\text{B}\sin\text{x}+2\text{B}\cos\text{x}$
$\Rightarrow8\cos\text{x}+\sin\text{x}-(3\text{A}+2\text{B})\cos\text{x}+(2\text{A}-3\text{B})\sin\text{x}$
Equating the coefficient of like terms we get,
$2\text{A}-3\text{B}=1\ \dots(2)$
$3\text{A}+2\text{B}=8\ \dots(3)$
Solving eq (2) and eq (3) we get,
$\text{A}=2,\text{B}=1$
Thus, by substracting the value of A and B in eq (1) we get,
$\text{I}=\int\Big[\frac{2(3\cos\text{x}+2\sin\text{x})+1(-3\sin\text{x}+2\cos\text{x})}{(3\cos\text{x}+2\sin\text{x})}\Big]\text{dx}$
$=2\int\Big(\frac{3\cos\text{x}+2\sin\text{x}}{3\cos\text{x}+2\sin\text{x}}\Big)\text{dx}+\int\Big(\frac{-3\sin\text{x}+2\cos\text{x}}{3\cos\text{x}+2\sin\text{x}}\Big)\text{dx}$
$=2\int\text{dx}+\int\Big(\frac{-3\sin\text{x}+2\cos\text{x}}{3\cos\text{x}+2\sin\text{x}}\Big)\text{dx}$
Putting $3\cos\text{x}+2\sin\text{x}=\text{t}$
$\Rightarrow(-3\sin\text{x}+2\cos\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=2\int\text{dx}+\int\frac{1}{\text{t}}\text{dt}$
$=2\text{x}+\ln|\text{t}|+\text{C}$
$=2\text{x}+\ln|3\cos\text{x}+2\sin\text{x}|+\text{C}$
View full question & answer→Question 3895 Marks
Evaluate the following integrals:
$\int\limits^{\text{a}}_{-\text{a}}\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)\text{d}\theta$
AnswerLet $\text{I}=\int\limits^{\text{a}}_{-\text{a}}\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)\text{d}\theta$
Consider $\text{f}(\theta)=\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)$
$\text{f}(-\theta)=\log\Big(\frac{\text{a}-\sin(-\theta)}{\text{a}+\sin(-\theta)}\Big)$
$=\log\Big(\frac{\text{a}+\sin\theta}{\text{a}-\sin\theta}\Big)$ $[\sin(-\text{x})=-\sin\text{x}\big]$
$=\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)^{-1}$
$=-\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)$ $\Big[\log\text{a}^{\text{b}}=\text{b}\log\text{a}\Big]$
$=-\text{f}(\theta)$
$\therefore\ \text{f}(-\theta)=-\text{f}(\theta)$
$\Rightarrow\text{I}=\int\limits^{\text{a}}_{-\text{a}}\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)\text{d}\theta=0$ $\begin{bmatrix}\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\begin{cases}2\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx},&\text{ if }\text{f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{f}(-\text{x})=-\text{f}(\text{x})\end{cases}\end{bmatrix}$
View full question & answer→Question 3905 Marks
Evaluate the following integrals:$\int\frac{\text{x}^2\tan^{-1}\text{x}}{1+\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\Big(\frac{\text{x}^2\tan^{-1}\text{x}}{1+\text{x}^2}\Big)\text{dx}$
$=\int\Big(\frac{\text{x}^2+1-1}{\text{x}^2+1}\Big)\tan^{-1}\text{x dx}$
$=\int\Big(1-\frac{1}{\text{x}^2+1}\Big)\tan^{-1}\text{x dx}$
$=\int1.\tan^{-1}\text{x dx}-\int\frac{\tan^{-1}\text{x}}{\text{x}^2+1}\text{dx}$
$=\Big[\tan^{-1}\text{x}\int1\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\text{x}\big)\int1\text{dx}\Big\}\text{dx}\Big]-\int\frac{\tan^{-1}\text{x}}{\text{x}^2+1}\text{dx}$
$=\Big[\tan^{-1}\text{x}\times\text{x}-\int\frac{\text{x}}{1+\text{x}^2}\text{dx}\Big]-\int\frac{\tan^{-1}\text{x}}{\text{x}^2+1}\text{dx}$
Putting $\text{x}^2+1=\text{t}$ in the first integral and $\tan^{-1}\text{x}=\text{p}$ in the second integral
$\Rightarrow2\text{x dx = dt} $ and $\frac{1}{1+\text{x}^2}\text{dx = dp}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$ and $\frac{1}{1+\text{x}^2}\text{dx = dp}$
$\therefore \text{I}=\tan^{-1}\text{x.x}-\frac{1}{2}\int\frac{\text{dt}}{\text{t}}-\int\text{p.dp}$
$=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\ln|\text{t}|-\frac{\text{P}^2}{2}+\text{C}$
$=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\ln|1+\text{x}^2|-\frac{(\tan^{-1}\text{x})^2}{2}+\text{C}$ $[\therefore \text{t}=\text{x}^2+1\text{ and}\text{ p}=\tan^{-1}\text{x}]$
View full question & answer→Question 3915 Marks
Evaluate the following:
$\int\limits^{\frac{\pi}{2}}_0\frac{\tan\text{x dx}}{1+\text{m}^2\tan^2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\tan\text{x dx}}{1+\text{m}^2\tan^2\text{x}}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\frac{\sin\text{x}}{\cos\text{x}}}{1+\text{m}^2\cdot\frac{\sin^2\text{x}}{\cos^2\text{x}}}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin\text{x}\cos\text{x dx}}{1-\sin^2\text{x}+\text{m}^2\sin^2\text{x}}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin\text{x}\cos\text{x}}{1+(\text{m}^2-1)\sin^2\text{x}}\text{dx}$
Put $\sin^2\text{x}=\text{t}$ $\Rightarrow\ 2\sin\text{x}\cos\text{x}\text{ dx}=\text{dt}$
When $\text{x}\rightarrow0,$ then $\text{t}\rightarrow0$ and $\text{x}\rightarrow\frac{\pi}{2},$ then $\text{t}\rightarrow1$
$\therefore\ \text{I}=\frac{1}{2}\int\limits^1_0\frac{\text{dt}}{1+(\text{m}^2-1)\text{t}}$
$=\frac{1}{2}\Big[\frac{1}{\text{m}^2-1}\log(1+(\text{m}^2-1)\text{t})\Big]^1_0$
$=\frac{1}{2(\text{m}^2-1)}\big[\log(1+(\text{m}^2-1))-\log1\big]$
$=\frac{\log\text{m}^2}{2(\text{m}^2-1)}=\frac{2\log|\text{m}|}{2(\text{m}^2-1)}=\frac{\log|\text{m}|}{\text{m}^2-1}$
View full question & answer→Question 3925 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^2_{1}\big(\text{x}^2-1\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=1,\text{ b}=2,\text{ f(x)}=\text{x}^2-1,\text{ h}=\frac{2-1}{\text{n}}=\frac{1}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_{1}\big(\text{x}^2-1\big)\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(1)+\text{f}(1+\text{h})+\ ....\ +\text{f}\big\{1+(\text{n}-1)\text{h}\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[(1-1)+(\text{h}^2-1)+\ ....\ +\big\{(\text{n}-1)^2\text{h}^2-1\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{n}-1+\text{h}^2\big\{1^2+2^2+3^2\ ....\ +(\text{n}-1)^2\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{n}-1+\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\bigg[\text{n}-1+\frac{(\text{n}-1)(2\text{n}-1)}{6\text{n}}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\bigg\{1-\frac{1}{\text{n}}+\frac{1}{6}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\bigg\}$
$=1+\frac{1}{3}$
$=\frac{4}{3}$
View full question & answer→Question 3935 Marks
Evaluate the following integrals:
$\int\frac{\sin^3\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$
AnswerLet I $=\int\frac{\sin^3\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$$\therefore\text{I}=\int\frac{\sin^2\text{x}\sin\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$
$\Rightarrow\text{I}=\int\frac{(1-\cos^2\text{x})}{\sqrt{\cos\text{x}}}\sin\text{x dx}\ ...(1)$
Let $\cos\text{x}=\text{t}$ then, $\text{d}(\cos\text{x})=\text{dt}$
$\Rightarrow-\sin\text{x dx}=\text{dt}$
$\Rightarrow\sin\text{x dx}=-\text{dt}$
Putting $\cos\text{x}=\text{t}$ and $\sin\text{x dx}=-\text{dt}$ in equation (1), we get
$\text{I}=\int\frac{(1-\text{t}^2)}{\sqrt{\text{t}}}\times-\text{dt}$
$=\int\frac{\text{t}^2-1}{\sqrt{\text{t}}}\text{dt}$
$=\int\Bigg(\frac{\text{t}^2}{\text{t}^\frac{1}{2}}-\frac{1}{\text{t}^\frac{1}{2}}\Bigg)\text{dt}$
$=\int\Big(\text{t}^{2-\frac{1}{2}}-\text{t}^\frac{-1}{2}\Big)\text{dt}$
$=\int\Big(\text{t}^\frac{3}{2}-\text{t}^{\frac{-1}{2}}\Big)\text{dt}$
$=\frac{2}{5}\text{t}^\frac{5}{2}-2\text{t}^\frac{1}{2}+\text{C}$
$\therefore\text{I}=\frac{2}{5}\cos^\frac{5}{2}\text{x}-2\cos^\frac{1}{2}\text{x}+\text{C}$
$\therefore\text{I}=\frac{2}{5}\cos^\frac{5}{2}\text{x}-2\sqrt{\cos\text{x}}+\text{C}$
View full question & answer→Question 3945 Marks
Evaluate the following:
$\int\limits^\pi_0\text{x}\sin\text{x}\cos^2\text{xdx}$
AnswerLet $\text{I}=\int\limits^\pi_0\text{x}\sin\text{x}\cos^2\text{xdx}\ \ \dots(\text{i})$
$\Rightarrow\ \text{I}=\int\limits^\pi_0(\pi-\text{x})\sin(\pi-\text{x})\cos^2(\pi-\text{x})\text{dx}$
$\bigg[\because\int\limits^\text{b}_\text{a}\text{f(x)}\text{dx}=\int\limits^\text{b}_\text{a}\text{f(a}+\text{b}-\text{a})\text{dx}\bigg]$
$\Rightarrow\ \text{I}=\int\limits^\pi_0(\pi-\text{x})\sin\text{x}\cos^2\text{x}\text{ dx}$
Adding Eqs. (i) and (ii), we get
$2\text{I}=\int\limits^\pi_0\pi\sin\text{x}\cos^2\text{x dx}$
Now, put $\cos\text{x}=\text{t}$
$\Rightarrow-\sin\text{x dx}=\text{dt}$
When $\text{x}\rightarrow0,$ then $\text{t}\rightarrow1$
and $\text{x}\rightarrow\pi,$ then $\text{t}\rightarrow-1$
$\therefore\ 2\text{I}=-\pi\int\limits^{-1}_1\text{t}^2\text{dt}$
$\Rightarrow\ 2\text{I}=-\pi\Big[\frac{\text{t}^3}{3}\Big]^{-1}_1$
$\Rightarrow\ 2\text{I}=-\frac{\pi}{3}[-1-1]=\frac{2\pi}{3}$
$\therefore\ \text{I}=\frac{\pi}{3}$
View full question & answer→Question 3955 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2+1}{(2\text{x}+1)(\text{x}^2-1)}\text{ dx}$
AnswerWe have,$\text{I}=\int\frac{\text{x}^2+1}{(2\text{x}+1)(\text{x}^2-1)}$
$\text{I}=\int\frac{(\text{x}^2+1)\text{dx}}{(2\text{x}+1)(\text{x}^2-1)(\text{x}+1)}$
Let $\text{I}=\int\frac{(\text{x}^2+1)}{(2\text{x}+1)(\text{x}^2-1)(\text{x}+1)}=\frac{\text{A}}{2\text{x}+1}+\frac{\text{B}}{\text{x}-1}+\frac{\text{C}}{\text{x}+1}$
$\Rightarrow\int\frac{(\text{x}^2+1)}{(2\text{x}+1)(\text{x}^2-1)(\text{x}+1)}=\frac{\text{A}(\text{x}^2-1)+\text{B}(2\text{x}+1)(\text{x}+1)+\text{C}(2\text{x}+1)(\text{x}-1)}{(2\text{x}-1)(\text{x}-1)(\text{x}+1)}$
$\Rightarrow\text{x}^2+1=\text{A}(\text{x}^2-1)+\text{B}(2\text{x}+1)(\text{x}+1)\\+\text{C}(2\text{x}+1)(\text{x}-1)$
Putting x - 1 = 0
⇒ x = 1
1 + 1 = A × 0 + B × 0 + C (-2 + 1) (-1 - 1)
⇒ 2 = B (3) (2)
$\Rightarrow\text{B}=\frac{1}{3}$
Putting x + 1 = 0
⇒ x = -1
1 + 1 = A × 0 + B (-2 + 1)(-1 - 1)
⇒ 2 = C (-1) (-2)
⇒ C = 1
Putting 2x + 1 = 0
$\Rightarrow\text{x}=-\frac{1}{2}$
$\Big(-\frac{1}{2}\Big)^2+1=\text{A}\Big(\frac{1}{4}-1\Big)$
$\Rightarrow\frac{1}{4}+1=\text{A}\Big(-\frac{3}{4}\Big)$
$\Rightarrow\frac{5}{4}=\text{A}\Big(-\frac{3}{4}\Big)$
$\text{A}=-\frac{5}{3}$
$\therefore\text{I}=-\frac{5}{3}\int\frac{\text{dx}}{2\text{x}+1}+\frac{1}{3}\int\frac{\text{dx}}{\text{x}-1}+\int\frac{\text{dx}}{\text{x}+1}$
$=-\frac{5}{3}\times\frac{\log|2\text{x}+1|}{2}+\frac{1}{3}\log|\text{x}-1|+\log|\text{x}+1|+\text{C}$
$=-\frac{5}{6}\log|2\text{x}-1|+\frac{1}{3}\log|\text{x}-1|+\log|\text{x}+1|+\text{C}$
View full question & answer→Question 3965 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^{\frac{\pi}{2}}_{0}\cos\text{x dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=\frac{\pi}{2},\text{ f(x)}=\cos\text{x},\text{ h}=\frac{\frac{\pi}{2}-0}{\text{n}}=\frac{\pi}{2\text{n}}$
Therefore, $\text{I}=\int\limits^{\frac{\pi}{2}}_{0}\cos\text{x dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big(0+(\text{n}-1)\text{h}\big)\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\cos0+\cos\text{h}+\cos2\text{h}+\ ....+\ \cos(\text{n}-1)\text{h}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Bigg[\frac{\cos\big((\text{n}-1)\frac{\text{h}}{2}\big)\sin\frac{\text{nh}}{2}}{\sin\frac{\text{h}}{2}}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg[\frac{\cos\big(\frac{\pi}{4}-\frac{\text{h}}{2}\Big)\sin\frac{\pi}{4}}{\sin\frac{\text{h}}{2}}\Bigg]$ $\Big(\text{Using, nh}=\frac{\pi}{2}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg[\frac{\frac{\text{h}}{2}}{\sin\frac{\text{h}}{2}}\times2\cos\Big(\frac{\pi}{4}-\frac{\text{h}}{2}\Big)\sin\frac{\pi}{4}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\text{h}}{2}}{\sin\frac{\text{h}}{2}}\times\lim\limits_{\text{h}\rightarrow0}2\cos\Big(\frac{\pi}{4}-\frac{\text{h}}{2}\Big)\sin\frac{\pi}{4}$
$=2\cos\frac{\pi}{4}\sin\frac{\pi}{4}$
$=2\times\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}$
$=1$
View full question & answer→Question 3975 Marks
Integrate the function in Exercise:
$\text{x}\ \cos^{-1}\text{x}$
AnswerLet $\text{I}=\int\text{x}\cos^{-1}\text{x dx}$
Taking $\cos^{-1}\text{x}$ as first function and x as second function and integrating by parts, we obtain.
$=\cos^{-1}\text{x}\int\text{x} \ \text{dx}-\int\Bigg\{\Big(\frac{\text{d}}{\text{dx}}\cos^{-1}\text{x}\Big)\int\text{x} \ \text{dx}\Bigg\}\text{dx}$
$=\cos^{-1}\text{x}.\frac{\text{x}^2}{2}-\int\frac{-1}{\sqrt{1-\text{x}^2}}.\frac{\text{x}^2}{2}\text{dx}$
$=\frac{\text{x}^2\cos^{-1}\text{x}}{2}-\frac{1}{2}\int\frac{1-\text{x}^2-1}{\sqrt{1-\text{x}^2}}\text{dx}$
$=\frac{\text{x}^2\cos^{-1}\text{x}}{2}-\frac{1}{2}\int\Bigg\{\sqrt{1-\text{x}^2}+\Bigg(\frac{-1}{\sqrt{1-\text{x}^2}}\Bigg)\Bigg\}\text{dx}$
$=\frac{\text{x}^2\cos^{-1}\text{x}}{2}-\frac{1}{2}\int\sqrt{1-\text{x}^2}\text{dx}-\frac{1}{2}\int\Bigg(\frac{-1}{\sqrt{1-\text{x}^2}}\Bigg)\text{dx}$
$=\frac{\text{x}^2\cos^{-1}\text{x}}{2}-\frac{1}{2}\text{I}_1-\frac{1}{2}\cos^{-1}\text{x}\dots(\text{i})$
Where, $\text{I}_1=\int\sqrt{1-\text{x}^2}\text{dx}$
$\Rightarrow\ \text{I}_1=\text{x}\sqrt{1-\text{x}^2}-\int\frac{\text{d}}{\text{dx}}\sqrt{1-\text{x}^2}\int\text{x dx}$
$\Rightarrow\ \text{I}_1=\text{x}\sqrt{1-\text{x}^2}-\int\frac{-2\text{x}}{2\sqrt{1-\text{x}^2}}. \ \text{x dx}$
$\Rightarrow\ \text{I}_1=\text{x}\sqrt{1-\text{x}^2}-\int\frac{-\text{x}^2}{\sqrt{1-\text{x}^2}}.\text{dx}$
$\Rightarrow\ \text{I}_1=\text{x}\sqrt{1-\text{x}^2}-\int\frac{1-\text{x}^2-1}{\sqrt{1-\text{x}^2}}.\text{dx}$
$\Rightarrow\ \text{I}_1=\text{x}\sqrt{1-\text{x}^2}-\Bigg\{\int\sqrt{1-\text{x}^2}\text{dx}+\int\frac{-\text{dx}}{\sqrt{1-\text{x}^2}}\Bigg\}$
$\Rightarrow\ \text{I}_1=\text{x}\sqrt{1-\text{x}^2}-\{\text{I}_1+\cos^{-1}\text{x}\}$
$\Rightarrow\ 2\text{I}_1=\text{x}\sqrt{1-\text{x}^2}-\cos^{-1}\text{x}$
$\therefore \ \text{I}_1=\frac{\text{x}}{2}\sqrt{1-\text{x}^2}-\frac{1}{2}\cos^{-1}\text{x}$
Substituting in (i), we obtain
$\text{I}_1=\frac{\text{x}\cos^{-1}\text{x}}{2}-\frac{1}{2}\Bigg(\frac{\text{x}}{2}\sqrt{1-\text{x}^2}-\frac{1}{2}\cos^{-1}\text{x}\Bigg)-\frac{1}{2}\cos^{-1}\text{x}$
$=\frac{(2\text{x}^2-1)}{4}\cos^{-1}\text{x}-\frac{\text{x}}{4}\sqrt{1-\text{x}^2}+\text{C}$
View full question & answer→Question 3985 Marks
Evaluate the following integrals:
$\int\limits^{\pi}_0\frac{\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx},0<\alpha<\pi$
AnswerWe have,
$\text{I}=\int\limits^{\pi}_0\frac{\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\pi}_0\frac{\pi-\text{x}}{1+\cos\alpha\sin(\pi-\text{x})}\text{ dx}$
$=\int\limits^{\pi}_0\frac{\pi-\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get,
$2\text{I}=\int\limits^{\pi}_0\frac{\text{x}+\pi-\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\frac{\pi}{2}\int\limits^{\pi}_0\frac{1}{1+\cos\alpha\sin\text{x}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\frac{1}{1+\cos\alpha\sin\text{x}}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\frac{1}{1+\cos\alpha\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\frac{1+\tan^{2}\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}+2\cos\alpha\tan\frac{\text{x}}{2}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\frac{\sec^{2}\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}+2\cos\alpha\tan\frac{\text{x}}{2}}\text{ dx}$
Putting $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\text{x dx}=\text{dt}$
When $\text{x}\rightarrow0;\text{t}\rightarrow0$
and $\text{x}\rightarrow\pi;\text{t}\rightarrow\infty$
$\therefore\ \text{I}=\frac{\pi}{2}\int\limits^{\infty}_0\frac{2}{1+\text{t}^2+2\cos\alpha\text{t}}\text{ dt}$
$=\frac{\pi}{2}\int\limits^{\infty}_0\frac{2}{(\text{t}+\cos\alpha)^2-\cos^2\alpha+1}\text{ dt}$
$={\pi}\int\limits^{\infty}_0\frac{1}{(\text{t}+\cos\alpha)^2+\sin^2\alpha}\text{ dt}$
$=\pi\bigg[\frac{1}{\sin\alpha}\tan^{-1}\Big(\frac{1+\cos\alpha}{\sin\alpha}\Big)\bigg]^1_0$
$=\frac{\pi}{\sin\alpha}\Big[\tan^{-1}(\infty)-\tan^{-1}(\cot\alpha)\Big]$
$=\frac{\pi}{\sin\alpha}\bigg[\frac{\pi}{2}-\tan^{-1}\Big(\tan\Big(\frac{\pi}{2}-\alpha\Big)\Big)\bigg]$
$=\frac{\pi\alpha}{\sin\alpha}$
View full question & answer→Question 3995 Marks
Evaluate the following integrals:
$\int\limits^{\pi}_2\log(1-\cos\text{x})\text{dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_2\log(1-\cos\text{x})\text{dx}$
$=\int\limits^{\pi}_2\Big(2\sin^2\frac{\text{x}}{2}\Big)\text{dx}$
$=\int\limits^{\pi}_2\log2\text{ dx}+\int\limits^{\pi}_2\log\sin\frac{\text{x}}{2}\text{ dx}$
Let $\text{t}=\frac{\text{x}}{2}$ in these cong integral then $\text{dt}=\frac{1}{2}\text{ dx}$
When $\text{x}\rightarrow0;\text{t}\rightarrow0$ and $\text{x}\rightarrow\pi;\text{t}\rightarrow\frac{\pi}{2}$
$\text{I}=\log2\big[\text{x}\big]^{\pi}_0+4\int\limits^{\frac{\pi}{2}}_0\log\sin\text{t dt}$
$=\pi\log2+4\times\Big(-\frac{\pi}{2}\log2\Big)$ $\Bigg[\text{Where,}\int\limits^{\frac{\pi}{2}}_0\log\sin\text{t dt}=-\frac{\pi}{2}\log2\Bigg]$
$=-\pi\log2$
View full question & answer→Question 4005 Marks
Evaluate the following integrals:
$\int(\text{x}-3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
AnswerConsider the integral $\text{I}=\int(\text{x}-3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$Let us express $\text{x}-3=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+3\text{x}-18)+\mu$
$\Rightarrow\text{x}-3=\lambda(2\text{x}+3)+\mu$
$\Rightarrow\text{x}-3=2\lambda\text{x}+3\lambda+\mu$
Comparing the co-efficients, we have,
$2\lambda=1\text{ and }3\lambda+\mu=-3$
$\Rightarrow\lambda=\frac{1}{2}\text{ and }3\times\frac{1}{2}+\mu=-3$
$\Rightarrow\lambda=\frac{1}{2}\text{ and }\mu-3-\frac{3}{2}$
$\Rightarrow\lambda=\frac{1}{2}\text{ and }\mu=-\frac{9}{2}$
Then
$\text{x}-3=\lambda(2\text{x}+3)+\mu$
Now the integral $\text{I}=\int(\text{x}-3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
$=\int\Big(\frac{1}{2}(2\text{x}+3)-\frac{9}{2}\Big)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
$\text{I}=\frac{1}{2}\int(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}\\-\frac{9}{2}\int\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
$\Rightarrow\text{I}=\text{I}_1+\text{I}_2$
where, $\text{I}_1=\frac{1}{2}\int(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
and $\text{I}_2=-\frac{9}{2}\int\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
Let us consider the integral, $I_1$:
$\text{I}_1=\frac{1}{2}\int(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
Substituting, $\text{x}^2+3\text{x}-18=\text{t}$
$\Rightarrow(2\text{x}+3)\text{dx = dt}$
Thus,
$\text{I}_1=\frac{1}{2}\int\sqrt{\text{t}}\text{dt}$
$=\frac{1}{2}\times\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\text{C}$
$=\frac{1}{2}\times\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}+\text{C}$
$=\frac{1}{2}\times\frac{2}{3}\times\text{t}^{\frac{3}{2}}+\text{C}$
$=\frac{1}{3}\times\text{t}^{\frac{3}{2}}+\text{C}$
$=\frac{1}{3}\times(\text{x}^2+3\text{x}-18)^{\frac{3}{2}}+\text{C}$
Now consider the integral
$\text{I}_2=-\frac{9}{2}\int\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\text{x}^2+2\times\frac{3}{2}\text{x}+\Big(\frac{3}{2}\Big)^2-\Big(\frac{3}{2}\Big)^2-18}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\frac{9}{4}-18}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9}{4}+18\Big)}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9+72}{4}\Big)}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{81}{4}\Big)}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9}{2}\Big)^2}\text{dx}$
We know that,
$\int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{x}^2-\text{a}^2}-\frac{1}{2}\text{a}^2\log\Big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\Big|+\text{C}$
$\therefore\ \text{I}_2=-\frac{9}{2}\begin{Bmatrix}\frac{1}{2}\Big(\text{x}+\frac{3}{2}\Big)\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9}{2}\Big)^2}\\-\frac{1}{2}\Big(\frac{9}{2}\Big)^2\log\Big|\Big(\text{x}+\frac{3}{2}\Big)+\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9}{2}\Big)^2}\Big|\end{Bmatrix}+\text{C}$
$=-\frac{9}{4}\begin{Bmatrix}\Big(\frac{2\text{x}+3}{2}\Big)\sqrt{\text{x}^2+3\text{x}-18}-\Big(\frac{729}{4}\Big)\\\log\Big|\Big(\text{x}+\frac{3}{2}\Big)+\sqrt{\text{x}^2+3\text{x}-18}\Big|\end{Bmatrix}+\text{C}$
$=-\frac{9}{8}(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}+\frac{729}{16}\\\log\Big|\Big(\text{x}+\frac{3}{2}\Big)+\sqrt{\text{x}^2+3\text{x}-18}\Big|+\text{C}$
Thus,
$\text{I}=-\frac{1}{3}(\text{x}^2+3\text{x}-18)^{\frac{3}{2}}-\frac{9}{8}(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}\\+\frac{729}{16}\log\Big|\Big(\text{x}+\frac{3}{2}\Big)+\sqrt{\text{x}^2+3\text{x}-18}\Big|+\text{C}$
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