Question 1015 Marks
Evaluate the following integrals:
$\int\frac{1}{(\text{x}^2+2\text{x}+10)^2}\text{ dx}$
AnswerLet $\int\frac{1}{(\text{x}^2+2\text{x}+10)^2}\text{ dx}$
$=\int\frac{1}{\big[(\text{x}+1)^2+3^2\big]}\text{ dx}$
Let $\text{x}+1=3\tan\theta$
On differentiating both sides, we get
$\text{dx}=3\sec^2\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\frac{1}{\big[3^2\tan^2\theta+3^2\big]^2}3\sec^2\theta\text{ d}\theta$
$=\frac{1}{27}\int\frac{\sec^2\theta}{\sec^4\theta}\text{ d}\theta$
$=\frac{1}{27}\int\frac{1}{\sec^2\theta}\text{ d}\theta$
$=\frac{1}{27}\int\cos^2\theta\text{ d}\theta$
$=\frac{1}{54}\int(1+\cos2\theta)\text{d}\theta$
$=\frac{1}{54}\Big(\theta+\frac{\sin2\theta}{2}\Big)+\text{ C}$
$=\frac{1}{54}\Big(\theta+\frac{\tan\theta}{1+\tan^2\theta}\Big)+\text{C}$
$=\frac{1}{54}\begin{pmatrix}\tan^{-1}\frac{\text{x}+1}{3}+\frac{\tan\Big(\tan^{-1}\frac{\text{x}+1}{3}\Big)}{1+\tan^{2}\Big(\tan^{-1}\frac{\text{x}+1}{3}\Big)}\end{pmatrix}+\text{C}$
$=\frac{1}{54}\begin{pmatrix}\tan^{-1}\frac{\text{x}+1}{3}+\frac{\frac{\text{x}+1}{3}}{1+\Big(\frac{\text{x}+1}{3}\Big)^2}\end{pmatrix}+\text{C}$
$=\frac{1}{54}\Bigg(\tan^{-1}\frac{\text{x}+1}{3}+\frac{\frac{\text{x}+1}{3}}{\frac{\text{x}^2+2\text{x}+10}{9}}\Bigg)+\text{C}$
$=\frac{1}{54}\bigg(\tan^{-1}\frac{\text{x}+1}{3}+\frac{3(\text{x}+1)}{\text{x}^2+2\text{x}+10}\bigg)+\text{C}$
View full question & answer→Question 1025 Marks
Evaluate the following integrals:$\int(\text{x}+1)\text{e}^{\text{x}}\log(\text{xe}^{\text{x}})\text{dx}$
Answer$\int(\text{x}+1)\text{e}^{\text{x}}.\log(\text{xe}^{\text{x}})\text{dx}$ Let $\text{x e}^{\text{x}}=\text{t}$ $\Rightarrow\big(\text{x.e}^{\text{x}}+1.\text{e}^{\text{x}}\big)\text{dx=dt}$ $\therefore\int(\text{x}+1)\text{e}^{\text{x}}.\log(\text{x e}^\text{x})\text{dx}=\int1.\log (\text{t})\text{dt}$ $=\log\text{t}\int1\text{dt}-\int\big\{\frac{\text{d}}{\text{dt}}(\log\text{t)}-\int1\text{dt}\Big\}\text{dt}$ $=\log(\text{t})\times\text{t}-\int\frac{1}{\text{t}}\times\text{t dt}$ $=\text{t}\log(\text{t})-\text{t}+\text{C} \dots(1)$Substituting the value of t in eq (1)
$\Rightarrow\int(\text{x}+1)\text{e}^{\text{x}}.\log(\text{x e}^{\text{x}})\text{dx} = (\text{x e}^{\text{x}}).\log(\text{x e}^{\text{x}})-\text{x e}^{\text{x}}+\text{C}$ $=\text{x e}^{\text{x}}\Big\{\log(\text{x e}^{\text{x}})-1\Big\}+\text{C}$
View full question & answer→Question 1035 Marks
Evaluate the integral in Exercise:
$\int\limits^{1}_{0}\sin^{-1}\bigg(\frac{2\text{x}}{1+\text{x}^{2}}\bigg)\text{dx}$
Answer$\text{Let}\text{I}=\int\limits_{0}^{1}\sin^{-1}\bigg(\frac{2\text{x}}{1+\text{x}^{2}}\bigg)\text{dx}$
$\text{put}\ \text{x}=\tan\theta\ \text{so that}\ \text{dx}=\sec^{2}\theta\ \text{d}\theta\ \text{when}\ \text{x}=0,\tan\theta=0\Rightarrow\theta=0$
$\text{when}\ \text{x}=1,\tan\theta=1\ \Rightarrow\theta=\frac{\pi}{4}$
$\therefore |=\int^{\frac{\pi}{4}}_{0}\sin^{-1}\bigg(\frac{2\tan\theta}{1+\tan^{2}\theta}\bigg).\sec^{2}\theta\ \text{d}\theta$
$ =\int^{\frac{\pi}{4}}_{0}\sin^{-1}(\sin2\theta).\sec^{2}\theta\ \text{d}\theta=\int^{\frac{\pi}{4}}_{0}2\theta.\sec^{2}\theta\ \text{d}\theta=2\int^{\frac{\pi}{4}}_{0}\theta.\sec^{2}\theta\ \text{d}\theta$
$=2\left\{[\theta\tan\theta]^{\frac{\pi}{4}}_{1}-\int^{\frac{\pi}{4}}_{0}1.\tan\theta\ \text{d}\theta\right\}=2\left\{[\theta\tan\theta]^{\frac{\pi}{4}}_{0}+[\log\cos\theta]^{\frac{\pi}{4}}_{0}\right\}$
$=2\left\{\frac{\pi}{4}\tan\frac{\pi}{4}-0+\log\cos\frac{\pi}{4}-\log\cos0\right\}=2\bigg[\frac{\pi}{4}\times1-0+\log\frac{1}{\sqrt{2}}-\log1\bigg]$
$=2\bigg[\frac{\pi}{4}+\log1-\log\sqrt{2}\bigg]=\frac{\pi}{2}-\log2$
View full question & answer→Question 1045 Marks
Evalute the following integrals:
$\int\frac{\sin2\text{x}}{\text{a}^2+\text{b}^2\sin^2\text{x}}\text{dx}$
AnswerLet $\int\frac{\sin2\text{x}}{\text{a}^2+\text{b}^2\sin^2\text{x}}\text{dx}\ .....(\text{i})$
Let $\text{a}^2+\text{b}^2\sin^2\text{x}=\text{t}$ then,
$\text{d}\big(\text{a}^2+\text{b}^2\sin^2\text{x}\big)=\text{dt}$
$=\text{b}^2(2\sin\text{x}\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{\text{b}^2(2\sin\text{x}\cos\text{x})}$
$=\frac{\text{dt}}{\text{b}^2\sin2\text{x}}$
Putting $\text{a}^2+\text{b}^2\sin^2\text{x}=\text{t}$ and $\text{dx}=\frac{\text{dt}}{\text{b}^2\sin2\text{x}}$ in equation (i), we get,
$\text{I}=\int\frac{\sin2\text{x}}{\text{t}}\times\frac{\text{dt}}{\text{b}^2\sin2\text{x}}$
$=\frac{1}{\text{b}^2}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{\text{b}^2}\log|\text{t}|+\text{C}$
$=\frac{1}{\text{b}^2}\log|\text{a}^2+\text{b}^2\sin^2\text{x}|+\text{C}$
$\Rightarrow\text{I}=\frac{1}{\text{b}^2}\log|\text{a}^2+\text{b}^2\sin^2\text{x}|+\text{C}$
View full question & answer→Question 1055 Marks
Evaluate the following integrals:
$\int(\text{e}^{\log\text{x}}+\sin\text{x})\cos\text{x dx}$
AnswerLet $\text{I}=\int(\text{e}^{\log\text{x}}+\sin\text{x})\cos\text{x dx}$
$=\int(\text{x}+\sin\text{x})\cos\text{x dx}$
$=\int\text{x}\cos\text{x dx}+\int\sin\text{x}\cos\text{x dx}$
$=\big[\text{x}\int\cos\text{x dx}-\int(1\int\cos\text{x dx})\text{dx}\big]+\frac{1}{2}\int\sin2\text{x dx}$
$=\big[\text{x}\sin\text{x}-\int\sin\text{x dx}\big]+\frac{1}{2}\Big(-\frac{\cos2\text{x}}{2}\Big)+\text{C}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}-\frac{1}{4}\cos2\text{x}+\text{C}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}-\frac{1}{4}\big[1-2\sin^2\text{x}\big]+\text{C}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}-\frac{1}{4}+\frac{1}{2}\sin^2\text{x}+\text{C}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}+\frac{1}{2}\sin^2\text{x}+\text{C}-\frac{1}{4}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}+\frac{1}{2}\sin^2\text{x}+\text{k},$ where $\text{k}=\text{C}-\frac{1}{4}$
View full question & answer→Question 1065 Marks
Evaluate the following integrals:
$\int^\limits{\text{a}}_{-\text{a}}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}}\text{ dx}$
AnswerLet $\text{x}=\text{a}\cos2\theta$
Differentiating w.r.t. x, we get
$\text{dx}=-2\text{a}\sin2\theta$
Now, $\text{x}=-\text{a}\Rightarrow\theta=\frac{\pi}{2}$
$\text{x}=\text{a}\Rightarrow\theta=0$
$\therefore\ \int^\limits{\text{a}}_{-\text{a}}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}}\text{ dx}=\int^\limits0_\frac{\pi}{2}\sqrt{\frac{\text{a}(1-\cos2\theta)}{\text{a}\big(1+\cos2\theta)}}(-2\sin2\theta\big)\text{d}\theta$
$=2\text{a}\int^\limits{\frac{\pi}{\text{2}}}_{0}\frac{\sin\theta}{\cos\theta}\cdot\sin2\theta\text{ d}\theta$ $\begin{bmatrix}\because1-\cos2\theta=2\sin^2\theta\\1+\cos2\theta=2\cos^2\theta\\-\int^\limits\text{b}_\text{a}\text{f(x)}\text{dx}=\int^\limits\text{a}_\text{b}\text{f}(\text{x})\text{dx} \end{bmatrix}$
$=2\text{a}\int^\limits{\frac{\pi}{\text{2}}}_{0}\frac{\sin\theta\cdot2\sin\theta\cos\theta}{\cos\theta}$
$=4\text{a}\int^\limits{\frac{\pi}{\text{2}}}_{0}\sin^{2}\theta\text{ d}\theta$
$=2\text{a}\int^\limits{\frac{\pi}{\text{2}}}_{0}\big(1-\cos2\theta\big)\text{d}\theta$
$=2\text{a}\Big[\theta-\frac{\sin2\theta}{2}\Big]^{\frac{\pi}{2}}_0$
$=2\text{a}\Big[\frac{\pi}{2}-0-0+0\Big]$
$=\pi\text{a}$
View full question & answer→Question 1075 Marks
Evalute the following integrals:
$\int\frac{\sin2\text{x}}{\text{a}\cos^2\text{x}+\text{b}\sin^2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin2\text{x}}{\text{a}\cos^2\text{x}+\text{b}\sin^2\text{x}}\text{dx}$ $=\int\frac{\sin2\text{x}}{\text{a}(1-\sin^2\text{x})+\text{b}\sin^2\text{x}}\text{dx}$ $=\int\frac{\sin2\text{x}}{(\text{b}-\text{a})\sin^2\text{x}+\text{a}}\text{dx}$ Putting $\sin^2\text{x}=\text{t}$ $\Rightarrow2\sin\text{x}\cos\text{x}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\sin2\text{x}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\sin2\text{x }\text{dx}=\text{dt}$ $\therefore\text{I}=\int\frac{1}{(\text{b}-\text{a})\text{t}+\text{a}}\text{dt}$ $=\frac{1}{(\text{b}-\text{a})}\text{ln}|(\text{b}-\text{a})\text{t}+\text{a}|+\text{C}$ $\Big[\because\int\frac{1}{\text{ax}+\text{b}}\text{dx}=\frac{1}{\text{a}}\text{In}|\text{ax}+\text{b}|+\text{C}\Big]$ $=\frac{1}{(\text{b}-\text{a})}\text{ln}|(\text{b}-\text{a})\sin^2\text{x}+\text{a}|+\text{C}\ \big[\because\text{t}=\sin^2\text{x}\big]$ $=\frac{1}{(\text{b}-\text{a})}\text{ln}\big|\text{b}\sin^2\text{x}+\text{a}(1-\sin^2\text{x})\big|+\text{C}$$=\frac{1}{(\text{b}-\text{a})}\text{ln}\big|\text{b}\sin^2\text{x}+\text{a}\cos^2\text{x}\big|+\text{C}$
View full question & answer→Question 1085 Marks
Evalute the following integrals:
$\int\frac{\text{cosec x}}{\log\tan\frac{\text{x}}{2}}\text{dx}$
AnswerLet $\int\frac{\text{cosec x}}{\log\tan\frac{\text{x}}{2}}\text{dx}\ .....(\text{i})$
Let $\log\tan\frac{\text{x}}{2}=\text{t}$ then,
$\text{d}\Big[\log\tan\frac{\text{x}}{2}\Big]=\text{dt}$
$\Rightarrow\text{cosec x dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{\text{cosec x}}$
Putting $\log\tan\frac{\text{x}}{2}=\text{t}$ and $\text{dx}=\frac{\text{dt}}{\text{cosec x}}$ in equation (i), we get
$\text{I}=\int\frac{\text{cosec x}}{\text{t}}\times\frac{\text{dt}}{\text{cosec x}}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log\Big|\log\tan\frac{\text{x}}{2}\Big|+\text{C}$
$\therefore\text{I}=\log\Big|\log\tan\frac{\text{x}}{2}\Big|+\text{C}$
View full question & answer→Question 1095 Marks
Evaluate the following integrals:
$\int^\limits2_1\frac{1}{\text{x}(1+\log\text{x})^2}\text{ dx}$
AnswerLet $\text{I}=\int^\limits2_1\frac{1}{\text{x}(1+\log\text{x})^2}\text{ dx}$ Then,
Let $(1+\log\text{x})=\text{t}$ Then, $\frac{1}{\text{x}}\text{ dx}=\text{dt}$
When, $\text{x}=1,\text{t}=1$ and $\text{x}=2,\text{t}=1+\log2$
$\therefore\ \text{I}=\int^\limits{(1+\log2)}_{1}\frac{1}{\text{t}}\text{ dt}$
$\Rightarrow\text{I}=\Big[\frac{-1}{\text{t}}\Big]^{(1+\log2)}_1$
$\Rightarrow\text{I}=-\frac{1}{1+\log2}+1$
$\Rightarrow\text{I}=\frac{\log2}{1+\log2}$
View full question & answer→Question 1105 Marks
Evaluate the following integrals:
$\int\text{x}\Big(\frac{\sec2\text{x}-1}{\sec2\text{x}+1}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{x}\Big(\frac{\sec2\text{x}-1}{\sec2\text{x}+1}\Big)\text{dx}$
$=\int\text{x}\Big(\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\Big)\text{dx}$
$=\int\text{x}\Big(\frac{\sec^2\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$=\int\text{x}\tan^2\text{x dx}$
$=\int\text{x}(\sec^2\text{x}-1)\text{dx}$
$=\int\text{x}\sec^2\text{x dx}-\int\text{dx}$
$=\big[\text{x}\int\sec^2\text{x dx}-\int(1\int\sec^2\text{x dx})\text{dx}\big]-\frac{\text{x}^2}{2}$
$=\text{x}\tan\text{x}-\int\tan\text{x dx}-\frac{\text{x}^2}{2}$
$\text{I}=\text{x}\tan\text{x}-\log\sec\text{x}-\frac{\text{x}^2}{2}+\text{C}$
View full question & answer→Question 1115 Marks
Evaluate the following integrals:
$\int4\text{x}^3\sqrt{5-\text{x}^2}\text{ dx}$
Answer$\int4\text{x}^3\sqrt{5-\text{x}^2}\text{ dx}$
$=4\int\text{x}^2\times\text{x}\sqrt{5-\text{x}^2}\text{ dx}$
Let $5-\text{x}^2=\text{t}$
$\Rightarrow\text{x}^2=5-\text{t}$
$\Rightarrow2\text{x}=-\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{x dx}=-\frac{\text{dt}}{2}$
Now, $4\int\text{x}^2\times\text{x}\sqrt{5-\text{x}^2}\text{ dx}$
$=\frac{4}{-2}\int(5-\text{x})\sqrt{\text{t}}\text{ dt}$
$=-2\int5\text{t}^{\frac{1}{2}}+2\int\text{t}^{\frac{3}{2}}\text{ dt}$
$=-10\Bigg[\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\Bigg]+2\Bigg[\frac{\text{t}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\Bigg]+\text{C}$
$=-\frac{20}{3}\text{t}^{\frac{3}{2}}+\frac{4}{5}\text{t}^{\frac{5}{2}}+\text{C}$
$=-\frac{20}{3}\big(5-\text{x}^2\big)^{\frac{3}{2}}+\frac{4}{5}\big(5-\text{x}^2\big)^{\frac{5}{2}}+\text{C}$
View full question & answer→Question 1125 Marks
Find the integrals of the functions in Exercises:
$\frac{1}{\cos(\text{x}-\text{a})\cos(\text{x}-\text{b})}$
Answer$\frac{1}{\cos(\text{x}-\text{a})\cos(\text{x}-\text{b})}$
$=\frac{1}{\sin(\text{a}-\text{b})}\bigg[\frac{\sin(\text{a}-\text{b})}{\cos(\text{x}-\text{a})\cos(\text{x}-\text{b})}\bigg]$
$=\frac{1}{\sin(\text{a}-\text{b})}\Bigg[\frac{\sin\big[(\text{x}-\text{b})-(\text{x}-\text{a})\big]}{\cos(\text{x}-\text{a})\cos(\text{x}-\text{b})}\Bigg]$
$=\frac{1}{\sin(\text{a}-\text{b})}\frac{\big[\sin(\text{x}-\text{b})\cos(\text{x}-\text{a})-\cos(\text{x}-\text{b})\sin(\text{x}-\text{a})\big]}{\cos(\text{x}-\text{a})\cos(\text{x}-\text{b})}$
$=\frac{1}{\sin(\text{a}-\text{b})}\big[{\tan(\text{x}-\text{b})-\tan(\text{x}-\text{a})}\big]$
$\Rightarrow\int\frac{1}{\cos(\text{x}-\text{a})\cos(\text{x}-\text{b})}\text{ dx}=\frac{1}{\sin(\text{a}-\text{b})}\int\big[{\tan(\text{x}-\text{b})-\tan(\text{x}-\text{a})}\big]\text{dx}$
$=\frac{1}{\sin(\text{a}-\text{b})}\Big[-\log\big|\cos(\text{x}-\text{b})\big|+\log\big|\cos(\text{x}-\text{a})\big|\Big]$
$=\frac{1}{\sin(\text{a}-\text{b})}\Bigg[\log\bigg|\frac{\cos(\text{x}-\text{a})}{\cos(\text{x}-\text{b})}\bigg|\Bigg]+\text{C}$
View full question & answer→Question 1135 Marks
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}-\cos^3\text{x}}(\sec^3\text{x}-1)\cos^2\text{x dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}-\cos^3\text{x}}(\sec^3\text{x}-1)\cos^2\text{x dx}$
$=\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}(1-\cos^2\text{x}\big)}(-\tan^2\text{x})\cos^2\text{x dx}$
$=-\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}(\sin^2\text{x})}\sin^2\text{x dx}$
$=-\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}}(\sin\text{x})\sin^2\text{x dx}$
$=-\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}}\big(1-\cos^2\text{x}\big)\sin\text{x dx}$ $\Big(\sin\text{x}=\sin\text{x}\text{ for }0\leq\text{x}\leq\frac{\pi}{2}\Big)$
Put $\cos\text{x}=\text{z}^2$
$\therefore\ -\sin\text{x dx}=2\text{z dz}$
When $\text{x}\rightarrow0,\text{ z}\rightarrow0$
When $\text{x}\rightarrow\frac{\pi}{2},\text{ z}\rightarrow0$
$\therefore\ \text{I}=-\int^\limits0_1\text{z}(1-\text{z}^4)2\text{z dz}$
$=-2\int^0\limits_1\text{z}^2\text{ dz}+2\int^\limits0_1\text{z}^6\text{ dz}$
$=-2\times\Big[\frac{\text{z}^3}{3}\Big]^0_1+2\times\Big[\frac{\text{z}^7}{7}\Big]^0_1$
$=-\frac{2}{3}\big(0-1\big)+\frac{2}{7}\big(0-1\big)$
$=\frac{2}{3}-\frac{2}{74}$
$=\frac{8}{21}$
View full question & answer→Question 1145 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\sin^3\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sin^3\text{x}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x }\sin^2\text{x}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x}(1-\cos^2\text{x})\text{dx}$
Let $\text{u}=\cos\text{x},\text{ du}=-\sin\text{x dx}$
$\therefore\ \text{I}=\int-(1-\text{u}^2)\text{du}$
$\Rightarrow\text{I}=\Big[\frac{\text{u}^3}{3}-\text{u}\Big]$
$\Rightarrow\text{I}=\Big[\frac{\cos^3\text{x}}{3}-\cos\text{x}\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=0-\frac{1}{3}+1$
$\Rightarrow\text{I}=\frac{2}{3}$
View full question & answer→Question 1155 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{1}\sqrt{\text{x}(1-\text{x})}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\sqrt{\text{x}(1-\text{x})}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{1}\sqrt{\frac{1}{4}-\big(\text{x}-\frac{1}{2}\big)^2}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{2}\int_{0}^\limits{1}\sqrt{1-\frac{\big(\text{x}-\frac{1}{2}\big)^2}{\frac{1}{4}}}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{2}\int_{0}^\limits{1}\sqrt{1-\Bigg(\frac{\text{x}-\frac{1}{2}}{\frac{1}{2}}\Bigg)^2}\text{ dx}$
Let $\Bigg(\frac{\text{x}-\frac{1}{2}}{\frac{1}{2}}\Bigg)=\sin\text{u}$
$\Rightarrow2\text{dx}=\cos\text{u du}$
$\therefore\ \text{I}=\frac{1}{4}\int_{-\frac{\pi}{2}}^\limits{\frac{\pi}{2}}\sqrt{1-\sin^2\text{u}}\cos\text{u du}$
$\Rightarrow \text{I}=\frac{1}{4}\int_{-\frac{\pi}{2}}^\limits{\frac{\pi}{2}}\cos^2\text{u du}$
$\Rightarrow \text{I}=\frac{1}{4}\int_{-\frac{\pi}{2}}^\limits{\frac{\pi}{2}}\Big(\frac{\cos2\text{u}+1}{2}\Big)\text{du}$
$\Rightarrow \text{I}=\frac{1}{8}\Big[\frac{\sin2\text{u}}{2}+\text{u}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}$
$\Rightarrow \text{I}=\frac{1}{8}\Big[\frac{\pi}{2}+\frac{\pi}{2}\Big]$
$\Rightarrow\text{I}=\frac{\pi}{8}$
View full question & answer→Question 1165 Marks
Integrate the function in Exercise: $\frac{\sqrt{\text{x}^{2}+1}\big[\log\text{(x}^{2}+1)-2\log\text{x}\big]}{\text{x}^{4}}$
Answer$\frac{\sqrt{\text{x}^{2}+1}\big[\log\text{(x}^{2}+1)-2\log\text{x}\big]}{\text{x}^{4}}=\frac{\sqrt{\text{x}^{2}+1}}{\text{x}^{4}}\big[\log\text{(x}^{2}+1)-\log\text{x}^{2}\big]$
$=\frac{\sqrt{\text{x}^{2}+1}}{\text{x}^{4}}\bigg[\log\bigg(\frac{\text{x}^{2}+1}{\text{x}^{2}}\bigg)\bigg]$
$=\frac{\sqrt{\text{x}^{2}+1}}{\text{x}^{4}}\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)$
$=\frac{1}{\text{x}^{3}}\sqrt{1+\frac{\text{x}^{2}+1}{\text{x}^{2}}}\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)$
$=\frac{1}{\text{x}^{3}}\sqrt{1+\frac{1}{\text{x}^{2}}}\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)$
$\text{Let}\ 1+\frac{1}{\text{x}^{2}}=\text{t}\Rightarrow\frac{-2}{\text{x}^{3}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{1}{\text{x}^{3}}\sqrt{1+\frac{1}{\text{x}^{2}}}\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)\text{dx}$
$=-\frac{1}{2}\int\sqrt{\text{t}}\log\text{t dt}$
$=-\frac{1}{2}\int\text{t}^{\frac{1}{2}}.\log\text{t dt}$
Integrating by parts, we obtain
$\text{I}=-\frac{1}{2}\Bigg[\log\text{t}.\int\text{t}^{\frac{1}{2}}\text{dt}-\left\{\bigg(\frac{\text{d}}{\text{dt}}\log\text{t}\bigg)\int\text{t}^{\frac{1}{2}}\text{dt}\right\}\text{dt}\Bigg]$
$=-\frac{1}{2}\begin{bmatrix}\log\text{t}.\frac{\text{t}\frac{3}{2}}{\frac{3}{2}}-\int\frac{1}{\text{t}}.\frac{\text{t}\frac{3}{2}}{\frac{3}{2}}\text{dt} \\ \end{bmatrix}$
$=-\frac{1}{2}\Bigg[\frac{2}{3}\text{t}^{\frac{3}{2}}\log\text{t}-\frac{2}{3}\int\text{t}^{\frac{1}{2}}\text{dt}\Bigg]$
$=-\frac{1}{2}\Bigg[\frac{2}{3}\text{t}^{\frac{3}{2}}\log\text{t}-\frac{4}{9}\int\text{t}^{\frac{3}{2}}\text{dt}\Bigg]$
$=-\frac{1}{3}\text{t}^{\frac{3}{2}}\log\text{t}+\frac{2}{9}\text{t}^{\frac{3}{2}}$
$=-\frac{1}{3}\text{t}^{\frac{3}{2}}\bigg[\log\text{t}-\frac{2}{3}\bigg]$
$=-\frac{1}{3}\bigg(1+\frac{1}{\text{x}^{2}}\bigg)^{\frac{3}{2}}\Bigg[\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)-\frac{2}{3}\Bigg]+\text{C}$
View full question & answer→Question 1175 Marks
Evaluate the following integrals:
$\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{3+\sin2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{3+\sin2\text{x}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{4-(1-\sin2\text{x})}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{4-(\sin^2\text{x}+\cos^2\text{x}-2\sin\text{x}\cos\text{x})}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{4-(\sin\text{x}-\cos\text{x})^2}\text{ dx}$
Put $\sin\text{x}-\cos\text{x}=\text{z}$
$\therefore\ (\cos\text{x}+\sin\text{x})\text{dx}=\text{dz}$
When $\text{x}\rightarrow0,\text{z}\rightarrow-1$ $(\text{z}=\sin0-\cos0=0-1=-1)$
When $\text{x}\rightarrow\frac{\pi}{4},\text{z}\rightarrow0$ $\Big(\text{z}=\sin\frac{\pi}{4}-\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0\Big)$
$\therefore\ \text{I}=\int_{-1}^\limits{0}\frac{\text{dz}}{2^2-\text{z}^2}$
$=\frac{1}{2\times2}\log\Big[\log\Big(\frac{2+\text{z}}{2-\text{z}}\Big)\Big]^0_{-1}$
$=\frac{1}{4}\Big(\log1-\log\frac{1}{3}\Big)$
$=\frac{1}{4}\big[0-\big(\log1-\log3\big)\big]$
$=-\frac{1}{4}\big(0-\log3\big)$
$=\frac{1}{4}\log3$
View full question & answer→Question 1185 Marks
Evaluate the following integrals:
$\int\frac{\sqrt{\tan\text{x}}}{\sin\text{x}\cos\text{x}}\text{dx}$
Answer$\int\frac{\sqrt{\tan\text{x}}}{\sin\text{x}\cos\text{x}}\text{dx}$
$=\int\frac{\sqrt{\tan\text{x}}}{\frac{\sin\text{x}}{\cos\text{x}}\times\cos^2\text{x}}\text{dx}$
$=\int\frac{\sqrt{\tan\text{x}}}{\tan\text{x}}\times\sec^2\text{x dx}$
$=\int\frac{1}{\sqrt{\tan\text{x}}}\times\sec^2\text{x dx}$
$=\int(\tan\text{x})^{-\frac{1}{2}}\sec^2\text{x dx}$
$\text{Let }\tan\text{x}=t$
$\Rightarrow\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec^2\text{x dx}=\text{dt}$
$\text{Now,}\int(\tan\text{x})^{-\frac{1}{2}}\sec^2\text{x dx}$
$=\int\text{t}^{{-\frac{1}{2}}}\text{dt}$
$=\Bigg[\frac{-{\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg]+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{\tan\text{x}}+\text{C}$
View full question & answer→Question 1195 Marks
Evaluate the following integrals:
$\int\sin^7\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\sin^7\text{x}\text{ dx}$ Then
$\text{I}=\int\sin^6\text{x }\sin\text{x}\text{ dx}$
$=\int\big(\sin^2\text{x}\big)^3\sin\text{x}\text{ dx}$
$=\int\big(1-\cos^2\text{x}\big)^3\sin\text{x}\text{ dx}$
$=\int\big(1-\cos^6\text{x}+3\cos^4\text{x}-3\cos^2\text{x}\big)\sin\text{x}\text{ dx}$
$\text{I}=\int\sin\text{x}\text{ dx}-\int\cos^6\text{x }\sin\text{x}\text{ dx}+3\int\cos^4\text{x }\sin\text{x}\text{ dx}-3\int\cos^2\text{x }\sin\text{x}\text{ dx}$
Putting $\cos\text{x}=\text{t}$ and $-\sin\text{x}\text{dx}=\text{dt}$ in $2^{nd}, 3^{rd}$ and $4^{th}$ integral we get
$\text{I}=\int\sin\text{x}\text{ dx}-\int\text{t}^6(-\text{dt})+3\int\text{t}^4(-\text{dt})-3\int\text{t}^2(-\text{dt})$
$=-\cos\text{x}+\frac{\text{t}^7}{7}-\frac{3}{5}\text{t}^5+\frac{3}{3}\text{t}^3+\text{C}$
$=-\cos\text{x}+\frac{\cos^7\text{x}}{7}-\frac{3}{5}\cos^5\text{x}+\cos^3\text{x}+\text{C}$
$\therefore\ \text{I}=-\cos\text{x}+\cos^3\text{x}-\frac{3}{5}\cos^5\text{x}+\frac{1}{7}\cos^7\text{x}+\text{C}$
View full question & answer→Question 1205 Marks
Evaluate the following integrals:$\int(\log\text{x})^2\cdot\text{x dx}$
AnswerLet $\text{I}=\int(\log\text{x})^2\text{x dx}$
Using integration by parts,
$=(\log\text{x})^2\int\text{x dx }-\int\Big(2(\log\text{x})\Big(\frac{1}{\text{x}}\Big)\int\text{x dx}\Big)\text{dx}$
$=\frac{\text{x}^2}{2}(\log\text{x})^2-2\int(\log\text{x})\Big(\frac{1}{\text{x}}\Big)\Big(\frac{\text{x}^2}{2}\Big)\text{dx}$
$=\frac{\text{x}^2}{2}(\log\text{x})^2-\int\text{x}(\log\text{x})\text{dx}$
$=\frac{\text{x}^2}{2}(\log\text{x})^2-\Big[\log\text{x}\int\text{x dx}-\int\Big(\frac{1}{\text{x}}\int\text{x dx}\Big)\text{dx}\Big]$
$=\frac{\text{x}^2}{2}(\log\text{x})^2-\Big[\frac{\text{x}^2}{2}\log\text{x}-\int\Big(\frac{1}{\text{x}}\times\frac{\text{x}^2}{2}\Big)\text{dx}\Big]$
$=\frac{\text{x}^2}{2}(\log\text{x})^2-\frac{\text{x}^2}{2}\log\text{x}+\frac{1}{2}\int\text{x dx}$
$=\frac{\text{x}^2}{2}(\log\text{x})^2-\frac{\text{x}^2}{2}\log\text{x}+\frac{1}{4}\text{x}^2+\text{C}$
$\text{I}=\frac{\text{x}^2}{2}\Big[(\log\text{x})^2-\log\text{x}+\frac{1}{2}\Big]+\text{C}$
View full question & answer→Question 1215 Marks
Evaluate the following integrals:
$\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2-\text{a}^2}}\text{ dx}$
Answer$\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2-\text{a}^2}}\text{ dx}$ Let $\text{x}^2=\text{t}$ $\Rightarrow2\text{x}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$ Now, $\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2-\text{a}^2}}\text{ dx}$$=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}+\text{a}^2}+\sqrt{\text{t}-\text{a}^2}}$
$=\frac{1}{2}\int\frac{\text{dt}}{\Big(\sqrt{\text{t}+\text{a}^2}+\sqrt{\text{t}-\text{a}^2}\Big)}\times\frac{\Big(\sqrt{\text{t}+\text{a}^2}-\sqrt{\text{t}-\text{a}^2}\Big)}{\Big(\sqrt{\text{t}+\text{a}^2}-\sqrt{\text{t}-\text{a}^2}\Big)}$
$=\frac{1}{2}\int\frac{\Big(\sqrt{\text{t}+\text{a}^2}-\sqrt{\text{t}-\text{a}^2}\Big)}{(\text{t}+\text{a}^2)-(\text{t}-\text{a}^2)}\text{ dt}$
$=\frac{1}{4\text{a}^2}\int\Big(\text{t}+\text{a}^2\Big)^{\frac12}\text{dt}-\frac{1}{4\text{a}^2}\big(\text{t}-\text{a}^2\big)^{\frac12}\text{dt}$
$=\frac{1}{4\text{a}^2}\begin{bmatrix}\frac{\big(\text{t}+\text{a}^2\big)^{\frac12+1}}{\frac{1}2+1}\end{bmatrix}-\frac{1}{4\text{a}^2}\begin{bmatrix}\frac{\big(\text{t}-\text{a}^2\big)^{\frac12+1}}{\frac12+1}\end{bmatrix}+\text{C}$
$=\frac{1}{6\text{a}^2}\begin{bmatrix}(\text{t}+\text{a}^2)^{\frac{3}{2}}-(\text{t}-\text{a}^2)^{\frac32}\end{bmatrix}+\text{C}$
$=\frac{1}{6\text{a}^2}\begin{bmatrix}(\text{x}^2+\text{a}^2)^{\frac32}-(\text{x}^2-\text{a}^2)^{\frac32}\end{bmatrix}+\text{C}$
View full question & answer→Question 1225 Marks
Evaluate the following definite integral as limit of sum:
$\int_\limits{0}^{5} \ \text{(x}+1) \ \text{d}\text{x}$
Answer$\text{we}\ \ \text{know}\ \text{that}\int_\limits{\text{a}}^\text{b}\ \text{f} \ \text{(x)} \ \text{dx}=\lim_\limits{\text{h}\rightarrow0} \text{h}[\text{f}\ \text{(a)}+\text{f}\ \text{(a}+\text{h)}+\text{f}\text{(a}+2\text{h)}+.......+\text{f} \text{(a}+\text{(n}-1)\text{h})]$
$\text{where}\ \text{nh}=\text{b}-\text{a}$
$\text{a}=0,\text{b}=5,\text{n}\text{h}= 5\ \text{and}\ \text{f}\text{(x)}=\text{x}+1$
$ \therefore \ \ \int_\limits{0}^{5}\text{(x}+1) \ \text{dx}=\lim_\limits{\text{h}\rightarrow0} \ \text{h}[1+\text{(h}+1)+(2\text{h}+1)+.......+\text{((n}-1)\text{h}+1)] $
$\Rightarrow\int_\limits{0}^{5}\text{(x}+1) \ \text{dx}=\lim_\limits{\text{h}\rightarrow0}\text{h}[\text{n}+\text{h}(1+2+3+.......+\text{(n}-1)]$
$\Rightarrow\int_\limits{0}^{5}\text{(x}+1) \ \text{dx}=\lim_\limits{\text{h}\rightarrow0}\Bigg[\text{n}\text{h}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Bigg]=\lim_\limits{\text{h}\rightarrow0}\Bigg[\text{n}\text{h}+\frac{\text{(nh}-\text{h)}}{2}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[5+\frac{5(5-\text{h)}}{2}\bigg]=\bigg[5+\frac{5(5-0)}{2}\bigg]=5+\frac{25}{2}=\frac{35}{2}$
View full question & answer→Question 1235 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}{\sqrt{\sin\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}+\sqrt{\cos\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}\text{ dx}$ $\Bigg[\int\limits^{\text{b}}_{\text{a}}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\big(\frac{\pi}{2}-\text{x}\big)}}{\sqrt{\sin\big(\frac{\pi}{2}-\text{x}\big)}+\sqrt{\cos\big(\frac{\pi}{2}-\text{x}\big)}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\cos\text{x}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\text{dx}$
$\Rightarrow2\text{I}=\Big[\text{x}\Big]^{\frac{\pi}{3}}_{\frac{\pi}{6}}$
$\Rightarrow2\text{I}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$
$\Rightarrow\text{I}=\frac{\pi}{12}$
View full question & answer→Question 1245 Marks
Evaluate the following definite integrals:
$\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}(\tan\text{x}+\cot\text{x})^2\text{ dx}$
AnswerWe have,
$\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}(\tan\text{x}+\cot\text{x})^2\text{ dx}$
$=\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\Big(\frac{\sin^2\text{x}+\cot^2\text{x}}{\sin\text{x}\cos\text{x}}\Big)^2\text{dx}$
$=\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\Big(\frac{1}{\sin\text{x}\cos\text{x}}\Big)^2\text{dx}$
Multiplying numberator and denominator by 2
$=\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\Big(\frac{2}{2\sin\text{x}\cos\text{x}}\Big)^2\text{dx}$
$=\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\Big(\frac{2}{\sin2\text{x}}\Big)^2\text{dx}$ $\big[\because2\sin\text{x}\cos\text{x}=\sin2\text{x}\big]$
$=4\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\text{cosec}^2\text{x dx}$
$=4\Big[-\frac{\cot2\text{x}}{2}\Big]^{\frac{\pi}{4}}_\frac{\pi}{3}$
$=2\Big[-\cot\frac{\pi}{2}+\cot2\frac{\pi}{3}\Big]$
$=2\Big[\frac{-1}{\sqrt{3}}-0\Big]$
$=\frac{-2}{\sqrt{3}}$
$\therefore\ \int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}(\tan\text{x}+\cot\text{x})^2\text{ dx}=\frac{-2}{\sqrt{3}}$
View full question & answer→Question 1255 Marks
Evalute the following integrals:
$\int\frac{-\sin\text{x}+2\cos\text{x}}{2\sin\text{x}+\cos\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{-\sin\text{x}+2\cos\text{x}}{2\sin\text{x}+\cos\text{x}}\text{dx}\ .....\text{(i)}$
Let $2\sin\text{x}+\cos\text{x}=\text{t}$ then,
$\text{d}(2\sin\text{x}+\cos\text{x})=\text{dt}$
$\Rightarrow(2\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{-\sin\text{x}+2\cos\text{x}}$
Putting $2\sin\text{x}+\cos\text{x}=\text{t and dx}=\frac{\text{dt}}{-\sin\text{x}+2\cos\text{x}}$ in equation (i), we get,
$\text{I}=\int\frac{-\sin\text{x}+2\cos\text{x}}{\text{t}}\times\frac{\text{dt}}{-\sin\text{x}+2\cos\text{x}}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|2\sin\text{x}+\cos\text{x}|+\text{C}$
$\therefore\text{I}=\log|2\sin\text{x}+\cos\text{x}|+\text{C}$
View full question & answer→Question 1265 Marks
Integrate the rational function in exercise:
$\frac{1}{(\text{e}^\text{x}-1)}$
$[$Hint: Put $e^x = t]$
Answer$\text{I}=\int\frac{1}{\text{e}^\text{x}-1}\text{dx}\dots(\text{i})$ Putting $e^x = t$
$\Rightarrow \ \text{e}^\text{x}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow \text{e}^{\text{x}} \text{dx = dt} \Rightarrow \text{x}$$\Rightarrow \ \text{dx}=\frac{\text{dt}}{\text{e}^\text{x}}$
$\therefore$ From eq. (i), $\text{I}=\int\frac{1}{\text{t}-1}\frac{\text{dt}}{\text{e}^\text{x}}=\int\frac{1}{\text{t}-1}\frac{\text{dt}}{\text{t}}=\int\frac{1}{\text{t}(\text{t}-1)}\text{dt}=\int\frac{\text{t}-(\text{t}-1)}{\text{t}(\text{t}-1)}\text{dt}$ $=\int\Bigg(\frac{\text{t}}{\text{t}(\text{t}-1)}-\frac{(\text{t}-1)}{\text{t}(\text{t}-1)}\Bigg)\text{dt}=\int\Bigg(\frac{1}{(\text{t}-1)}-\frac{1}{\text{t}}\Bigg)\text{dt}=\int\frac{1}{\text{t}-1}\text{dt}-\int\frac{1}{\text{t}}\text{dt}$$=\text{log}|\text{t}-1|-\text{log}|\text{t}|+\text{c}=\text{log}\Bigg|\frac{\text{t}-1}{\text{t}}\Bigg|+\text{c}=\text{log}\Bigg|\frac{\text{e}^\text{x}-1}{\text{e}^\text{x}}\Bigg|+\text{c}$
View full question & answer→Question 1275 Marks
Evaluate the following integrals:
$\int\limits^2_{-2}\text{xe}^{|\text{x}|}\text{ dx}$
Answer$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}+\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}$
For
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}$
Using integration by parts
$\int\text{f}^\text{t}\text{g}=\text{fg}-\int\text{fg}^{\text{t}}$
$\text{f}^\text{t}=\text{e}^{-\text{x}},\text{g}=\text{x}$
$\text{f}=-\text{e}^{-\text{x}},\text{g}^\text{t}=1$
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}=\big\{-\text{xe}^{-\text{x}}\big\}^0_{-2}+\int\limits^0_{-2}\text{e}^{-\text{x}}\text{ dx}$
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}=\big\{-\text{xe}^{-\text{x}}-\text{e}^{-\text{x}}\big\}^0_{-2}$
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}=\big\{(-1)-\big(2\text{e}^2-\text{e}^2\big)\big\}$
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}=\big\{-1-\text{e}^2\big\}$
For
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}$
Using integration by parts
$\int\text{f}^\text{t}\text{g}=\text{fg}-\int\text{fg}^{\text{t}}$
$\text{f}^\text{t}=\text{e}^{\text{x}},\text{g}=\text{x}$
$\text{f}=-\text{e}^{-\text{x}},\text{g}^\text{t}=1$
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}=\big\{\text{xe}^{\text{x}}\big\}^2_{0}-\int\limits^2_{0}\text{e}^{\text{x}}\text{ dx}$
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}=\big\{\text{xe}^{\text{x}}-\text{e}^{\text{x}}\big\}^2_{0}$
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}=2\text{e}^2-\text{e}^2+1$
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}=\text{e}^2+1$
Hence answer is,
$\int\limits^2_{-2}\text{xe}^{|\text{x}|}\text{ dx}=-1-\text{e}^2+\text{e}^2+1=0$
View full question & answer→Question 1285 Marks
Evaluate the following integrals:
$\int_\limits{\frac{1}{3}}^{1}\frac{\big(\text{x}-\text{x}^3\big)^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
AnswerLet $\text{I}=\int_\limits{\frac{1}{3}}^{1}\frac{\big(\text{x}-\text{x}^3\big)^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
$=\int_\limits{\frac{1}{3}}^{1}\frac{\Bigg[\text{x}^3\Big(\frac{\text{x}}{\text{x}^3}-1\Big)\Bigg]^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
$=\int_\limits{\frac{1}{3}}^{1}\frac{\text{x}\big(\frac{1}{\text{x}^2}-1\big)^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
$=\int_\limits{\frac{1}{3}}^{1}\frac{\text{x}\big(\frac{1}{\text{x}^2}-1\big)^{\frac{1}{3}}}{\text{x}^3}\text{ dx}$
Put $\Big(\frac{1}{\text{x}^2}-1\Big)=\text{Z}$
$\therefore\ -\frac{2}{\text{x}^3}\text{ dx}=\text{dz}$
$\Rightarrow\frac{\text{dx}}{\text{x}^3}=-\frac{\text{dz}}{2}$
When $\text{x}\rightarrow\frac{1}{3},\text{z}\rightarrow8$
When $\text{x}\rightarrow1,\text{z}\rightarrow0$
$\therefore\ \text{I}=-\frac{1}{2}\int^\limits0_8\text{z}^{\frac{1}{3}}\text{ dz}$
$=-\frac{1}{2}\times\Bigg[\frac{\text{z}^{\frac{4}{3}}}{\frac{4}{3}}\Bigg]^0_8$
$=-\frac{3}{8}\Big[0-(8)^{\frac{4}{3}}\Big]$
$=-\frac{3}{8}\times(-16)$
$=6$
View full question & answer→Question 1295 Marks
Evalute the following integrals:
$\int\frac{1-\cot\text{x}}{1+\cot\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1-\cot\text{x}}{1+\cot\text{x}}\text{dx}$ then,
$\text{I}=\int\frac{1-\frac{\cos\text{x}}{\sin\text{x}}}{1+\frac{\cos\text{x}}{\sin\text{x}}}\text{dx}$
$=\int\frac{\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}}}{\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}}}\text{dx}$
$\Rightarrow\text{I}=\int\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{dx}\ .....(\text{i})$
Let $\sin\text{x}+\cos\text{x}=\text{t},$ then,
$\text{d}(\sin\text{x}+\cos\text{x})=\text{dt}$
$\Rightarrow(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\Rightarrow-(\sin\text{x}-\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=-\frac{\text{dt}}{\sin\text{x}-\cos\text{x}}$
Putting $\sin\text{x}+\cos\text{x}=\text{t and dx}=-\frac{\text{dt}}{\sin\text{x}-\cos\text{x}}$ in equation (i), we het
$\text{I}=\int\frac{\sin\text{x}-\cos\text{x}}{\text{t}}\times\frac{-\text{dt}}{\sin\text{x}-\cos\text{x}}$
$=\int\frac{-\text{dt}}{\text{t}}$
$=-\log|\text{t}|+\text{C}$
$=-\log|\sin\text{x}+\cos\text{x}|+\text{C}$
View full question & answer→Question 1305 Marks
Evalute the following integrals:
$\int\tan2\text{x}\tan3\text{x}\tan5\text{x dx}$
AnswerLet $\text{I}=\int1+\tan\text{x}\tan(\text{x}+\theta)\text{dx}$
$=\int1+\tan\text{x}\Big(\frac{\tan\text{x}+\tan\theta}{1-\tan\text{x}\tan\theta}\Big)\text{dx}$
$=\int\frac{1+\tan^2\text{x}}{1-\tan\text{x}\tan\theta}\text{dx}$
$=\int\frac{\sec^2\text{x dx}}{1-\tan\text{x}\tan\theta}$
Putting $\tan\text{x}=\text{t}$
$\Rightarrow\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{\sec^2\text{x}}$
$\therefore\text{I}\approx\int\frac{1}{1-\text{t}\tan\theta}\text{dt}$
$=\frac{-1}{\tan\theta}\text{ ln}|1-\text{t}\tan\theta|+\text{C}$
$\Big[\because\int\frac{1}{\text{ax}+\text{b}}\text{dx}=\frac{1}{\text{a}}\text{ ln}|\text{ax}+\text{b}|+\text{C}\Big]$
$=-\cot\theta\text{ ln}|1-\tan\text{ x }\tan\theta|+\text{C}$
$=\cot\theta\text{ ln}\Big|\frac{1}{1-\tan\text{ x }\tan\theta}\Big|+\text{C}$
$=\cot\theta\text{ ln}\Big|\frac{\cos\text{ x }\cos\theta}{\cos\text{x}\cos\theta-\sin\text{x}\sin\theta}\Big|+\text{C}$
$=\cot\theta\text{ ln}\Big|\frac{\cos\text{x}}{\cos(\text{x}+\theta)}\Big|+\text{C }\big[\text{Let C}'=\text{C}+\cot\theta\text{ ln}\cos\theta\big]$
View full question & answer→Question 1315 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\text{dx}}{1+\cot\big(\frac{\pi}{2}-\text{x}\big)}\text{dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\text{dx}}{1+\tan\text{x}}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot\text{x}}+\frac{1}{1+\tan\text{x}}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1+\tan\text{x}+1+\cot\text{x}}{(1+\cot\text{x})(1+\tan\text{x})}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{2+\tan\text{x}+\cot\text{x}}{1+\tan\text{x}+\cot\text{x}+\tan\text{x}\cot\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{2+\tan\text{x}+\cot\text{x}}{2+\tan\text{x}+\cot\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$
View full question & answer→Question 1325 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\tan\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\tan\text{x}}}\text{ dx}\ ....(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\tan\big(\frac{\pi}{2}-\text{x}}\big)}\text{dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\cot\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\tan\text{x}}}+\frac{1}{1+\sqrt{\cot\text{x}}}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1+\sqrt{\cot\text{x}}+1+\sqrt{\tan\text{x}}}{\big(1+\sqrt{\tan\text{x}}\big)\big(1+\sqrt{\cot\text{x}}\big)}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1+\sqrt{\cot\text{x}}+1+\sqrt{\tan\text{x}}}{1+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}+\sqrt{\tan\text{x}\cot\text{x}}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0$
$=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$
View full question & answer→Question 1335 Marks
$\int\limits_{0}^{1}\text{x}\log(1+2\text{x})\text{dx}$
AnswerLet $\text{I}=\int\limits_{0}^{1}\text{x}\log(1+2\text{x})\text{dx}$
$=\Bigg[\log(1+2\text{x})\frac{\text{x}^2}{2}\Bigg]_{0}^{1}-\int\frac{1}{1+2\text{x}}\cdot2\cdot\frac{\text{x}^2}{2}\text{dx}$
$=\frac{1}{2}\big[\text{x}^2\log(1+2\text{x})\big]_{0}^{1}-\int\frac{\text{x}^2}{1+2\text{x}}\text{dx}$
$=\frac{1}{2}[\log3-0]-\Bigg[\int\limits_{0}^{1}\bigg(\frac{\text{x}}{2}-\frac{\frac{\text{x}}{2}}{1+2\text{x}}\bigg)\text{dx}\Bigg]$
$=\frac{1}{2}\log3-\frac{1}{2}\int\limits^{1}_{0}\text{x dx}+\frac{1}{2}\int\limits^{1}_{0}\frac{\text{x}}{1+2\text{x}}\text{dx}$
$=\frac{1}{2}\log3-\frac{1}{2}\bigg[\frac{\text{x}^2}{2}\bigg]^{1}_{0}+\frac{1}{2}\int\limits^{1}_{0}\frac{\frac{1}{2}(2\text{x}+1-1)}{(2\text{x}+1)}\text{dx}$
$=\frac{1}{2}\log3-\frac{1}{2}\Big[\frac{1}{2}-0\Big]+\frac{1}{4}\int^1_0\text{dx}-\frac{1}{4}\int^1_0\frac{1}{1+2\text{x}}\text{dx}$
$=\frac{1}{2}\log3-\frac{1}{4}+\frac{1}{4}\big[\text{x}\big]^{1}_{0}-\frac{1}{8}\big[\log|(1+2\text{x})|\big]^{1}_{0}$
$=\frac{1}{2}\log3-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}\big[\log3-\log1\big]$
$=\frac{1}{2}\log3-\frac{1}{8}\log3$
$=\frac{3}{8}\log3$
View full question & answer→Question 1345 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^3_{2}\big(2\text{x}^2+1\big)\text{ dx}$
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, $\text{a}=2,\text{ b}=3,\text{ f(x)}=2\text{x}^2+1,\text{ h}=\frac{3-2}{\text{n}}=\frac{1}{\text{n}}$
Therefore, $\text{I}=\int\limits^3_{2}\big(2\text{x}^2+1\big)\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(2)+\text{f}(2+\text{h})+\ ....\ +\text{f}\big\{2+(\text{n}-1)\text{h}\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2(2.2)^2+1+\big\{2(2+\text{h})^2+1\big\}+\\\ ....\ +\big\{2((2+\text{n}-1)\text{h})^2+1\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+2\Big\{2^2+(2+\text{h})^2+\ .....\big((2+\text{n}-1\big)\text{h}\big)^2\Big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+8\text{n}+2\text{h}^2\Big\{1^2+2^2+3^3+\ .....\ +(\text{n}-1)^2\Big\}\\+8\text{h}\big\{1+2+\ ....\ +(\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\text{h}\bigg[9\text{n}+\text{h}^2\frac{2\text{n}(\text{n}-1)(2\text{n}-1)}{6}+8\text{h}\frac{\text{n}(\text{n}-1)}{2}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\bigg[9\text{n}+\frac{(\text{n}-1)(2\text{n}-1)}{3\text{n}}+4\text{n}-4\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\bigg\{13+\frac{1}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{4}{\text{n}}\bigg\}$
$=13+\frac{2}{3}$
$=\frac{41}{3}$
View full question & answer→Question 1355 Marks
Evaluate the following integrals:
$\int_{0}^\limits{{\frac{\pi}{4}}}(\tan\text{x}+\cot\text{x})^{-2}\text{ dx}$
Answer$\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}(\tan\text{x}+\cot\text{x})^{-2}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\frac{1}{(\tan\text{x}+\cot\text{x})^{2}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\frac{1}{\Big(\frac{\sin^2\text{x}+\cos^2\text{x}}{\sin\text{x}\cos\text{x}}\Big)^2}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}(\sin\text{x}\cos\text{x})^2\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\sin^2\text{x}(1-\sin^2\text{x})\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\sin^2\text{x dx}-\int_{0}^\limits{{\frac{\pi}{4}}}\sin^4\text{x dx}$
We know that by reduction formula,
$\int\sin^{\text{n}}\text{x dx}=\frac{\text{n}-1}{\text{n}}\int\sin^{\text{n}-2}\text{x dx}-\frac{\cos\text{x}\sin^{\text{n}-1}}{\text{n}}$
For n = 2
$\Rightarrow\int\sin^2\text{x dx}=\frac{2-1}{2}\int1\text{ dx}-\frac{\cos\text{x}\sin\text{x}}{2}$
$\Rightarrow\int\sin^2\text{x dx}=\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}$
For n = 4
$\Rightarrow\int\sin^4\text{x dx}=\frac{4-1}{4}\int\sin^2\text{x dx}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
$\Rightarrow\int\sin^4\text{x dx}=\frac{3}{4}\Big\{\frac{1}{2}\times-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
Hence,
$\text{I}=\Big\{\frac{1}{2}\times-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}^{\frac{\pi}{4}}_0-\Big\{\frac{3}{4}\Big\{\frac{1}{2}\times-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}\Big\}^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\Big\{\frac{\pi}{8}-\frac{1}{4}\Big\}-\Big\{\frac{3}{4}\Big(\frac{\pi}{8}-\frac{1}{4}\Big)-\frac{1}{16}\Big\}$
$\Rightarrow\text{I}=\frac{\pi}{32}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}(\sin^{\text{x}}\cos\text{x})^2\text{dx}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}\sin^2\text{x}(1-\sin^2\text{x})\text{dx}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}\sin^2\text{x}-\sin^4\text{x dx}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}\sin^2\text{x dx}-\int_{0}^\limits{\frac{\pi}{2}}\sin^4\text{x dx}$
We know, by reduction formula,
$\int\sin^{\text{n}}\text{x dx}=\frac{\text{n}-1}{\text{n}}\int\sin^{\text{n}-2}\text{x dx}-\frac{\cos\text{x }\sin^{\text{n}-1}\text{x}}{\text{n}}$
For n = 2
$\Rightarrow\int\sin^2\text{x dx}=\frac{2-1}{2}\int1\text{ dx}-\frac{\cos\text{x}\sin\text{x}}{2}$
$\Rightarrow\int\sin^2\text{x dx}=\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}$
For n = 4
$\Rightarrow\int\sin^4\text{x dx}=\frac{4-1}{4}\int\sin^2\text{x dx}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
$\Rightarrow\int\sin^4\text{x dx}=\frac{3}{4}\Big\{\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
Hence,
$\Big\{\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}^{\frac{\pi}{2}}_0-\Big\{\frac{3}{4}\Big\{\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}\Big\}^{\frac{\pi}{2}}_0$
$\Rightarrow\frac{\pi}{4}-\frac{3}{4}\Big\{\frac{\pi}{4}\Big\}$
$\Rightarrow\frac{\pi}{16}$
View full question & answer→Question 1365 Marks
Evaluate the following integrals:
$\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^2\text{x dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^2\text{x dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\pi}_0(\pi-\text{x})\sin(\pi-\text{x})\cos^2(\pi-\text{x})\text{dx}$ $\Bigg[\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin\text{x}\cos^2\text{x dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\pi}_0(\pi-\text{x}+\text{x})\sin\text{x}\cos^2\text{x dx}$
$\Rightarrow2\text{I}=\pi\int\limits^{\pi}_0\sin\text{x}\cos^2\text{x dx}$
$\Rightarrow2\text{I}=-\pi\int\limits^{\pi}_0\cos^2\text{x}(-\sin\text{x})\text{dx}$
$\Rightarrow2\text{I}=-\pi\Big[\frac{\cos^3\text{x}}{3}\Big]^{\pi}_0$ $\Bigg[\int\big[\text{f(x)}\big]^{\text{n}}\text{f}'(\text{x})\text{dx}=\frac{\big[\text{f(x)}\big]^{\text{n}+1}}{\text{n}+1}+\text{C}\Bigg]$
$\Rightarrow2\text{I}=-\frac{\pi}{3}\big(\cos^3\text{x}-\cos^20\big)$
$\Rightarrow2\text{I}=-\frac{\pi}{3}(-1-1)=\frac{2\pi}{3}$
$\Rightarrow\text{I}=\frac{\pi}{3}$
View full question & answer→Question 1375 Marks
Evaluate the following integrals:
$\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}$
AnswerWe know,
$\int\limits^{2\pi}_0\text{f(x)}\text{dx}=\int\limits^{2\pi}_0\text{f}(2\pi-\text{x})\text{dx}$
Hence,
$\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}=\int\limits^{2\pi}_0\log\big(\sec(2\pi-\text{x})+\tan(2\pi-\text{x})\big)\text{dx}$
$\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}=\int\limits^{2\pi}_0\log(\sec\text{x}-\tan\text{x})\text{dx}$
If
$\text{I}=\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}+\int\limits^{2\pi}_0\log(\sec\text{x}-\tan\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}+\log(\sec\text{x}-\tan\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(\sec^2\text{x}-\tan^2\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(1)\text{dx}$
$2\text{I}=0$
$\text{I}=0$
View full question & answer→Question 1385 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^2(\text{x}^4+1)^{\frac{3}{4}}}\text{ dx}$
Answer$\int\frac{1}{\text{x}^2(\text{x}^4+1)^{\frac{3}{4}}}\text{ dx}$
Multiplying and dividing by $x^{-3},$ we obtain
$\frac{\text{x}^{-3}}{\text{x}^2.\text{x}^{-3}(\text{x}^4+1)^{\frac{3}{4}}}=\frac{\text{x}^{-3}(\text{x}^4+1)^{\frac{-3}{4}}}{\text{x}^2.\text{x}^{-3}}$
$=\frac{(\text{x}^4+1)^{\frac{-3}{4}}}{\text{x}^5.(\text{x}^{4})^{-\frac{3}{4}}}$
$=\frac{1}{\text{x}^5}\Big(\frac{\text{x}^4+1}{\text{x}^4}\Big)^{-\frac{3}{4}}$
$=\frac{1}{\text{x}^5}\Big(1+\frac{1}{\text{x}^4}\Big)^{-\frac{3}{4}}$
Let, $\frac{1}{\text{x}^4}=\text{t}$
$\Rightarrow-\frac{4}{\text{x}^5}\text{ dx}=\text{dt}$
$\Rightarrow\frac{1}{\text{x}^5}\text{ dx}=-\frac{\text{dt}}{4}$
View full question & answer→Question 1395 Marks
Evaluate the following integrals:
$\int\limits^\pi_0\sin^{100}\text{x}\cos^{101}\text{x dx}$
AnswerLet $\text{I}=\int\limits^\pi_0\sin^{100}\text{x}\cos^{101}\text{x dx}$
Consider $\text{f(x)}=\sin^{100}\text{x}\cos^{101}\text{x}$
Now,
$\text{f}(2\pi-\text{x})=\sin^{100}(2\pi-\text{x})\cos^{101}(2\pi-\text{x})$
$=(-\sin\text{x})^{100}(\cos\text{x})^{101}=\sin^{100}\text{x}\cos^{101}\text{x}=\text{f}(\text{x)}$
$\therefore\ \text{I}=\int\limits^{2\pi}_0\sin^{100}\text{x}\cos^{101}\text{x dx}=2\int\limits^{\pi}_0\sin^{100}\text{ x}\cos^{101}\text{x 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}$
Again,
$\text{f}(\pi-\text{x})=\sin^{100}(\pi-\text{x})\cos^{101}(\pi-\text{x})$
$=(\sin\text{x})^{100}(-\cos\text{x})^{101}=-\sin^{100}\text{x}\cos^{101}\text{x}=-\text{f(x)}$
$\therefore\ \text{I}=2\times0=0$ $\begin{bmatrix}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx},&\text{if }\text{f}(2\text{a}-\text{x})=\text{f(x)}\\0,&\text{if }\text{f}(2\text{a}-\text{x})=\text{f(x)}\end{cases}\end{bmatrix}$
View full question & answer→Question 1405 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\sin\text{x}}\text{ dx}$
AnswerLet $\int_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\sin\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\sin\text{x}}\times\frac{\sqrt{1-\sin\text{x}}}{\sqrt{1-\sin\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sqrt{1-\sin^2\text{x}}}{\sqrt{1-\sin^2\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\cos\text{x}}{\sqrt{1-\sin\text{x}}}\text{ dx}$
Let $1-\sin\text{x}=\text{u}$
$\Rightarrow-\cos\text{x dx}=\text{du}$
$\therefore\ \text{I}=\int\frac{-\text{du}}{\sqrt{\text{u}}}$
$\Rightarrow\text{I}=\big[-2\sqrt{\text{u}}\big]$
$\Rightarrow\text{I}=\big[-2\sqrt{1-\sin\text{x}}\big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=0+2$
$\Rightarrow\text{I}=2$
View full question & answer→Question 1415 Marks
Evaluate the definite integral in Exercise:
$\int^{\frac{\pi}{2}}_{0}\sin2\text{x}\tan^{-1}(\sin\text{x})\text{dx}$
Answer$\text{Let I}=\int^{\frac{\pi}{2}}_{0}\sin2\text{x}\tan^{-1}(\sin\text{x})\text{dx}=\int^{\frac{\pi}{2}}_{0}2\sin\text{x}\cos\text{x}\tan^{-1}(\sin\text{x})\text{dx} $
Also, let $\sin\text{x}=\text{t}=\ \Rightarrow\cos\text{x}\ \text{dx}=\text{dt}$
when $\text{x}=0,\text{t}=0$ and when $\text{x}=\frac{\pi}{2},=1$
$\Rightarrow \text{I}=2\int^{1}\limits_{0}\text{t}\cdot\tan^{-1}\text{t dt}$
Consider $\int\text{t}.\tan^{-1}\text{dt}=\tan^{-1}\text{t}.\int\text{t}\ \text{dt}-\int\left\{\frac{\text{d}}{\text{dt}}(\tan^{-1}\text{t)}\int\text{t}\ \text{dt}\right\}\text{dt}$
$=\tan^{-1}\text{t}.\frac{\text{t}^{2}}{2}-\int\frac{1}{1+\text{t}^{2}}.\frac{\text{t}^{2}}{2}\text{dt}$
$=\frac{\text{t}^{2}\tan^{-1}\text{t}}{2}-\frac{1}{2}\int\frac{\text{t}^{2}+1-1}{1+\text{t}^{2}}\text{dt}$
$=\frac{\text{t}^{2}\tan^{-1}\text{t}}{2}-\frac{1}{2}\int1\text{dt}+\frac{1}{2}\int\frac{1}{1+\text{t}^{2}}\text{dt}$
$=\frac{\text{t}^{2}\tan^{-1}\text{t}}{2}-\frac{1}{2}.\text{t}+\frac{1}{2}\tan^{-1}\text{t}$
$\Rightarrow\int^{1}\limits_{0}\text{t}.\tan^{-1}\text{t}\ \text{dt}=\bigg[\frac{\text{t}^{2}.\tan^{-1}\text{t}}{2}-\frac{\text{t}}{2}+\frac{1}{2}\tan^{-1}\text{t}\bigg]^{1}_{0}$
$=\frac{1}{2}\bigg[\frac{\pi}{4}-1+\frac{\pi}{4}\bigg]$
$=\frac{1}{2}\bigg[\frac{\pi}{4}-1\bigg]=\frac{\pi}{4}-\frac{1}{2}$
From equation (1), we obtain
$\text{I}=2\bigg[\frac{\pi}{4}-\frac{1}{2}\bigg]=\frac{\pi}{2}-1$
View full question & answer→Question 1425 Marks
Evaluate the following:
$\int\sqrt{\frac{\text{a}+\text{x}}{\text{a}-\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{\frac{\text{a}+\text{x}}{\text{a}-\text{x}}}\text{dx}$
Put $\text{x}=\text{a}\cos2\theta$ $\Rightarrow\ =\text{a}\cdot\sin2\theta\cdot2\cdot\text{d}\theta$
$\therefore\ =-2\int\sqrt{\frac{\text{a}+\text{a}\cos2\theta}{\text{a}-\text{a}\cos2\theta}}\cdot\text{a}\sin2\theta\text{ d}\theta$
$=-2\text{a}\int\sqrt{\frac{1+\cos2\theta}{1-\cos2\theta}}\sin2\theta\text{ d}\theta$
$=-2\int\sqrt{\frac{2\cos^2\theta}{2\sin^2\theta}}\sin2\theta\text{ d}\theta$
$=-2\text{a}\int\cot\theta\cdot\sin2\theta\text{ d}\theta$
$=-2\text{a}\int\frac{\cos\theta}{\sin\theta}\cdot2\sin\theta\cos\theta\text{ d}\theta$
$=-4\text{a}\int\cos^2\theta\text{ d}\theta$
$=-2\text{a}\int(1+\cos2\theta)\text{d}\theta$
$=-2\text{a}\Big[\theta+\frac{1}{2}\sin2\theta\Big]+\text{C}$
$=-2\text{a}\bigg[\frac{1}{2}\cos^{-1}\frac{\text{x}}{\text{a}}+\frac{1}{2}\sqrt{1-\frac{\text{x}^2}{\text{a}^2}}\bigg]+\text{C}$
$=-\text{a}\bigg[\cos^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\sqrt{1-\frac{\text{x}^2}{\text{a}^2}}\bigg]+\text{C}$
View full question & answer→Question 1435 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\frac{\tan^{-1}\text{x}}{1+\text{x}^2}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\frac{\tan^{-1}\text{x}}{1+\text{x}^2}\text{ dx}$ Then,
Let $\tan^{-1}\text{x}=\text{t}$ Then, $\frac{1}{1+\text{x}^2}\text{ dx}=\text{dt}$
When $\text{x}=0,\text{t}=0$ and $\text{x}=1,\text{t}=\frac{\pi}{4}$
$\therefore\ \text{I}=\int_{0}^\limits{\frac{\pi}{4}}\text{t}\text{ dt}$
$\Rightarrow\text{I}=\Big[\frac{\text{t}^2}{2}\Big]^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\frac{\pi^2}{32}$
View full question & answer→Question 1445 Marks
Integrate the following integrals:
$\int\sin\text{x}\cos2\text{x}\sin3\text{x dx}$
Answer$\int\sin\text{x}\cos2\text{x}\sin3\text{x dx}$
$=\frac{1}{2}\int(2\sin\text{x}\cos2\text{x})\sin3\text{x dx}$
$=\frac{1}{2}\int\big[\sin(\text{x}+2\text{x})+\sin(\text{x}-2\text{x})\big]\sin(3\text{x) dx}$
$=\frac{1}{2}\int\big[\sin(3\text{x})-\sin(\text{x})\big]\sin(3\text{x) dx}$
$=\frac{1}{2}\big[\int\sin^2(3\text{x})\text{dx}-\int\sin(\text{x})\sin(3\text{x})\text{dx}\big]$
$=\frac{1}{4}\big[\int2\sin^2(3\text{x})\text{dx}-\int2\sin(\text{x})\sin(3\text{x})\text{dx}\big]$
$=\frac{1}{4}\Big\{\int\big[1-\cos(6\text{x})\big]\text{dx}-\int\big[\cos(\text{x}-3\text{x})-\cos(\text{x}+3\text{x})\big]\text{dx}\Big\}$
$=\frac{1}{4}\big[\int1\text{dx}-\int\cos(6\text{x})\text{dx}-\int\cos(2\text{x})\text{dx}+\int\cos(4\text{x})\text{dx}\big]$
$=\frac{1}{4}\Big[\text{x}-\frac{\sin(6\text{x})}{6}-\frac{\sin(2\text{x})}{2}+\frac{\sin(4\text{x})}{4}\Big]+\text{C}$
$=\frac{\text{x}}{4}-\frac{\sin(6\text{x})}{24}-\frac{\sin(2\text{x})}{8}+\frac{\sin(4\text{x})}{16}+\text{C}$
Hence, $\int\sin\text{x}\cos2\text{x}\sin3\text{x}\text{ dx}$ $=\frac{\text{x}}{4}-\frac{\sin(6\text{x})}{24}-\frac{\sin(2\text{x})}{8}+\frac{\sin(4\text{x})}{16}+\text{C}$
View full question & answer→Question 1455 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\big(\text{a}^2\cos^2\text{x}+\text{b}^2\sin^2\text{x}\big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\big(\text{a}^2\cos^2\text{x}+\text{b}^2\sin^2\text{x}\big)\text{dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\big(\text{a}^2\cos^2\text{x}+\text{b}^2(1-\cos^2\text{x})\big)\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\big(\text{b}^2+(\text{a}^2-\text{b}^2)\cos^2\text{x}\big)\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\Bigg(\text{b}^2+\frac{\big(\text{a}^2-\text{b}^2\big)\big(1+\cos2\text{x}\big)}{2}\Bigg)\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\bigg[\text{b}^2\text{x}+\frac{\text{a}^2-\text{b}^2}{2}\Big(\text{x}+\frac{\sin2\text{x}}{2}\Big)\bigg]_0^{\frac{\pi}{2}}$
$\Rightarrow\text{I}=\frac{\text{b}^2\pi}{2}+\frac{\text{a}^2-\text{b}^2}{2}\frac{\pi}{2}+0$
$\Rightarrow\text{I}=\frac{\pi}{4}\big(\text{a}^2+\text{b}^2\big)$
View full question & answer→Question 1465 Marks
Evaluate the following intregals:
$\int\frac{5\text{x}}{(\text{x}+1)(\text{x}^2-4)}\text{dx}$
Answer$\int\frac{5\text{x}}{(\text{x}+1)(\text{x}^2-4)}=\frac{5\text{x}}{(\text{x}+1)(\text{x}+2)(\text{x}-2)}$
Let $\frac{5\text{x}}{(\text{x}+1)(\text{x}+2)(\text{x}-2)}=\frac{\text{A}}{(\text{x}+1)}+\frac{\text{B}}{(\text{x}+2)}+\frac{\text{C}}{(\text{x}-2)}$
$5\text{x}=\text{A}(\text{x}+2)(\text{x}-2)+\text{B}(\text{x}+1)(\text{x}-2)\\+\text{C}(\text{x}+1)(\text{x}+2)\ \dots(1)$
Substituting x = -1, -2 and 2 respectively in equation (1), we obtain
$\text{A}=\frac{5}{3},\text{B}=\frac{5}{2},\text{and }\text{C}=\frac{5}{6}$
$\therefore\frac{5\text{x}}{(\text{x}+1)(\text{x}+2)(\text{x}-2)}=\frac{5}{3(\text{x}+1)}-\frac{5}{2(\text{x}+2)}+\frac{5}{6(\text{x}-2)}$
$\Rightarrow\frac{5\text{x}}{(\text{x}+1)(\text{x}^2-4)}\ \text{dx}=\frac{5}{3}\int\frac{1}{(\text{x}+1)}\ \text{dx}-\frac{5}{2}\int\frac{1}{(\text{x}+2)}\\\ \text{dx}+\frac{5}{6}\int\frac{1}{(\text{x}-2)}\ \text{dx}$
$=\frac{5}{3}\log|\text{x}+1|-\frac{5}{2}\log|\text{x}+2|+\frac{5}{6}\log|\text{x}-2|+\text{C}$
View full question & answer→Question 1475 Marks
Evaluate the following integrals:
$\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\text{x}}{1-\text{x}}}\Big\}\text{dx}$
AnswerLet $\text{I}=\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\text{x}}{1-\text{x}}}\Big\}\text{dx}$
Let $\text{x}=\cos2\theta$
On differentiating both sides, we get
$\text{dx}=-2\sin2\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\cos2\theta}{1-\cos2\theta}}\Big\}2\sin2\theta\text{ d}\theta$
$=-2\int\cos\Big\{2\cot^{-1}\sqrt{\frac{2\cos^2\theta}{2\sin^2\theta}}\Big\}\sin2\theta\text{ d}\theta$
$=-2\int\cos\{2\cot^{-1}(\cot\theta)\}\sin2\theta\text{ d}\theta$
$=-2\int\cos2\theta\sin2\theta\text{ d}\theta$
$=\frac{\cos4\theta}{4}+\text{C}_1$
$=-\int\sin4\theta\text{ d}\theta$
$=\frac{1}{4}(2\cos^2\theta-1)+\text{C}_1$
$=\frac{1}{2}\text{x}^2-\frac{1}{4}+\text{C}_1$
$=\frac{1}{2}\text{x}^2+\text{C},$ where $\text{C}=-\frac{1}{4}+\text{C}_1$
Hence, $\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\text{x}}{1-\text{x}}}\Big\}\text{dx}=\frac{1}{2}\text{x}^2+\text{C}$
View full question & answer→Question 1485 Marks
Integrate the function in Exercise:
$\text{e}^{2\text{x}}\sin\text{x}$
AnswerLet $\text{I}=\int\text{e}^{2\text{x}}\sin\text{x dx}\dots(\text{I})$Integrating by parts, we obtain,
$\text{I}=\sin\text{x}\int\text{e}^{2\text{x}} \text{dx}-\int\Bigg\{\Bigg(\frac{\text{d}}{\text{dx}}\sin\text{x}\Bigg)\int\text{e}^{2\text{x}}\text{dx}\Bigg\}\text{dx}$
$\Rightarrow\ \text{I}=\sin\text{x}.\frac{\text{e}^{2\text{x}}} {2}-\int\cos\text{x}.\frac{\text{e}^{2\text{x}}}{2}\text{dx}$
$\Rightarrow\ \text{I}=\frac{\text{e}^{2\text{x}}\sin\text{x}} {2}-\frac{1}{2}\int\text{e}^{2\text{x}}\cos\text{x}\text{dx}$
Again integrating by parts, we obtain,
$\text{I}=\frac{\text{e}^{2\text{x}}.\sin\text{x}} {2}-\frac{1}{2}\Bigg[\cos\text{x}\int\text{e}^{2\text{x}}\text{dx}-\int\Bigg\{\Bigg(\frac{\text{d}}{\text{dx}}\cos\text{x}\Bigg)\int\text{e}^{2\text{x}}\text{dx}\Bigg\}\text{dx}\Bigg]$
$\Rightarrow\ \text{I}=\frac{\text{e}^{2\text{x}}.\sin\text{x}} {2}-\frac{1}{2}\Bigg[\cos\text{x}.\frac{\text{e}^{2\text{x}}}{2}-\int(-\sin\text{x})\frac{\text{e}^{2\text{x}}}{2}\text{dx}\Bigg]$
$\Rightarrow\ \text{I}=\frac{\text{e}^{2\text{x}}.\sin\text{x}} {2}-\frac{1}{2}\Bigg[\frac{\text{e}^{2\text{x}}\cos\text{x}}{2}+\frac{1}{2}\int\text{e}^{2\text{x}}\sin\text{x}\ \text{dx}\Bigg]$
$\Rightarrow\ \text{I}=\frac{\text{e}^{2\text{x}}.\sin\text{x}} {2}-\frac{\text{e}^{2\text{x}}\cos\text{x}}{4}-\frac{1}{4}\text{I}$ [From (I)]
$\Rightarrow\ \text{I}+\frac{1}{4}\text{I}=\frac{\text{e}^{2\text{x}}.\sin\text{x}} {2}-\frac{\text{e}^{2\text{x}}\cos\text{x}}{4}$
$\Rightarrow\ \frac{5}{4}\text{I}=\frac{\text{e}^{2\text{x}}.\sin\text{x}} {2}-\frac{\text{e}^{2\text{x}}\cos\text{x}}{4}$
$\Rightarrow\ \text{I}=\frac{4}{5}\Bigg[\frac{\text{e}^{2\text{x}}\sin\text{x}} {2}-\frac{\text{e}^{2\text{x}}\cos\text{x}}{4}\Bigg]+\text{C}$
$\Rightarrow\ \text{I}=\frac{\text{e}^{2\text{x}}}{5}\Bigg[2\sin\text{x}-\cos\text{x}\Bigg]+\text{C}$
View full question & answer→Question 1495 Marks
Evaluate the definite integral in Exercise: $\int^{\frac{\pi}{4}}_{0}\frac{\sin\text{x}+\cos\text{x}}{9+16\sin2\text{x}}\text{dx}$
Answer$\text{Let I}=\int^{\frac{\pi}{4}}_{0}\frac{\sin\text{x}+\cos\text{x}}{9+16\sin2\text{x}}\text{dx}$
Also, let $\sin\text{x}-\cos\text{x}=\text{t}\Rightarrow(\cos\text{x}+\sin\text{x})\text{dx}=\text{dt}$
when $\text{x}=0,\text{t}=-1$ and when $\text{x}=\frac{\pi}{4},\text{t}=0$
$\Rightarrow(\sin\text{x}-\cos\text{x})^{2}=\text{t}^{2}$
$\Rightarrow\sin^{2}\text{x}+\cos^{2}\text{x}-2\sin\text{x}\cos\text{x}=\text{t}^{2}$
$\Rightarrow1-\sin2\text{x}=\text{t}^{2}$
$\Rightarrow\sin2\text{x}=1-\text{t}^{2}$
$\therefore\text{I}=\int^{0}\limits_{-1}\frac{\text{dt}}{9+16(1-\text{t}^{2})}$
$=\int^{0}\limits_{-1}\frac{\text{dt}}{9+16-16\text{t}^{2}}$
$=\int^{0}\limits_{-1}\frac{\text{dt}}{25-16\text{t}^{2}}=\int^{0}\limits_{-1}\frac{\text{dt}}{(5)^{2}-(4\text{t})^{2}}$
$=\frac{1}{4}\Bigg[\frac{1}{2(5)}\log\Bigg|\frac{5+4\text{t}}{5-4\text{t}}\Bigg|\Bigg]^{0}_{-1}$
$=\frac{1}{40}\bigg[\log(1)-\log\bigg|\frac{1}{9}\bigg|\bigg]^{0}_{-1}$
$=\frac{1}{40}\log9$
View full question & answer→Question 1505 Marks
Integrate the function in Exercise:
$\frac{5\text{x}+3}{\sqrt{\text{x}^2+4\text{x}+10}}$
Answer$\text{Let I }=\int\frac{5\text{x}+3}{\sqrt{\text{x}^2+4\text{x}+10}}\text{ dx} \ \ \ \ \ ...\text{(i)}$
$\text{Let Linear}=\text{A}\frac{\text{d}}{\text{dx}}(\text{Quadratic})+\text{B}$
$\Rightarrow\ \ 5\text{x}+3=\text{A}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+4\text{x}+10\big)+\text{B}$
$\Rightarrow\ \ 5\text{x}+3=\text{A}(2\text{x}+4)+\text{B} \ \ \ \ ...\text{(ii)}$
$\Rightarrow\ \ 5\text{x}+3=2\text{A}\text{x}+4\text{A}+\text{B}$
Comparing coefficients of $x,$
$2\text{A}=5\ \ \Rightarrow\ \ \text{A}=\frac{5}{2}$
Comparing constants,
$4\text{A}+\text{B}=3$
On solving, we get
$\text{A}=\frac{5}{2}, \ \text{B}=-7$
Putting the values of $A$ and $B$ in eq. (ii),
$5\text{x}+3=\frac{5}{2}(2\text{x}+4)-7$
Putting this value of $5x + 3$ in eq. (i),
$\text{I}=\int\frac{\frac{5}{2}(2\text{x}+4)-7}{\sqrt{\text{x}^2+4\text{x}+10}}\text{ dx}$
$\text{I}=\frac{5}{2}\int\frac{2\text{x}+4}{\sqrt{\text{x}^2+4\text{x}+10}}\text{ dx}-7\int\frac{1}{\sqrt{\text{x}^2+4\text{x}+10}}\text{ dx}$
$\Rightarrow\ \ \text{I}=\frac{5}{2}\text{I}_1-7\ \text{I}_2\ \ \ \ ...\text{(iii)}$
$\text{Now I}_1=\int\frac{2\text{x}+4}{\sqrt{\text{x}^2+4\text{x}+10}}\text{ dx}$
$\text{Putting }\text{ x}^2+4\text{x}+10=\text{t}\ \ \Rightarrow\ \ \ 2\text{x}+4=\frac{\text{dt}}{\text{dx}}\ \ \Rightarrow\ \ \ (2\text{x}+4)\text{ dx}=\text{dt}$
$\therefore\ \ \ \text{I}_1=\int\frac{\text{dt}}{\sqrt{\text{t}}}=\int\text{t}^{\frac{-1}{2}}\text{ dt}=\frac{\text{t}^{\frac{1}{2}}}{\frac{1}{2}}$
$2\sqrt{\text{t}}=2\sqrt{\text{x}^2+4\text{x}+10} \ \ \ \ ...\text{(iv)}$
$\text{Again I}_2=\int\frac{1}{\sqrt{\text{x}^2+4\text{x}+10}}\text{ dx}$
$=\int\frac{1}{\sqrt{\text{x}^2+4\text{x}+4+6}}$
$=\int\frac{1}{\sqrt{(\text{x}+2)^2+\big(\sqrt{6}\big)^2}}\text{ dx}$
$=\log\begin{vmatrix}\text{x}+2+\sqrt{(\text{x}+2)^2+(6)^2}\end{vmatrix}$
$=\log\begin{vmatrix}\text{x}+2+\sqrt{\text{x}^2+4\text{x}+10}\end{vmatrix} \ \ \ \ ...\text{(v)}$
Putting values of $I_1$ and $I_2$ in eq. $(iii),$
$\text{I}=5\sqrt{\text{x}^2+4\text{x}+10}-7\log\begin{vmatrix}\text{x}+2+\sqrt{\text{x}^2+4\text{x}+10}\end{vmatrix}+\text{c}$
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