Question 513 Marks
Evaluate the following integrals:
$\int\frac{1+\cos\text{x}}{(\text{x}+\sin\text{x})^3}\text{dx}$
Answer$\int\frac{(1+\cos\text{x})}{(\text{x}+\sin\text{x})^3}\text{dx}$
$\text{Let x}+\sin\text{x}=\text{t}$
$\Rightarrow(1+\cos\text{x})=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(1+\cos\text{x})\text{dx}={\text{dt}}$
$\text{Now,}\int\frac{(1+\cos\text{x})}{(\text{x}+\sin\text{x})^3}\text{dx}$
$=\int\frac{\text{dt}}{\text{t}^3}$
$=\int\text{t}^{-3}\text{dt}$
$=\frac{\text{t}^{-3+1}}{-3+1}+\text{C}$
$=\frac{-1}{2\text{t}^2}+\text{C}$
$=\frac{-1}{2(\text{x}+\sin\text{x})^2}+\text{C}$
View full question & answer→Question 523 Marks
Evaluate the following intregals:
$\int\frac{1}{\sin^2\text{x}-\sin2\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sin^2\text{x}-\sin(2\text{x})}\ \text{dx}$
$=\int\frac{1}{\sin^2\text{x}+2\sin\text{x}\cos\text{x}}\text{ dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\sec^2\text{x}}{\tan\text{x}+2\tan\text{x}}\ \text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}^2+2\text{t}}$
$=\int\frac{\text{dt}}{\text{t}^2+2\text{t}+1-1}$
$=\ln\frac{\text{dt}}{(\text{t}+1)^2-(-1)^2}$
$=\frac{1}{2}\ln\Big|\frac{\text{t}}{\text{t}+2}\Big|+\text{C}$
$=\frac{1}{2}\ln\Big|\frac{\tan\text{x}}{\tan\text{x}+2}\Big|+\text{C}$
View full question & answer→Question 533 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{3+5\cos\text{x}}{3+5\sin\text{x}}\Big)\text{dx}$
AnswerLet, $\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{3+5\cos\text{x}}{3+5\sin\text{x}}\Big)\text{dx}\ ...(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\log\Bigg[\frac{3+5\cos\big(\frac{\pi}{2}-\text{x}\big)}{3+5\sin\big(\frac{\pi}{2}-\text{x}\big)}\Bigg]$
$=\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{3+5\sin\text{x}}{3+5\cos\text{x}}\Big)\text{dx}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\bigg[\log\Big(\frac{3+5\cos\text{x}}{3+5\sin\text{x}}\Big)+\log\Big(\frac{3+5\sin\text{x}}{3+5\cos\text{x}}\Big)\bigg]\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{3+5\cos\text{x}}{3+5\sin\text{x}}\times\frac{3+5\sin\text{x}}{3+5\cos\text{x}}\Big)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log1\text{ dx}=0$
Hence, $\text{I}=0$
View full question & answer→Question 543 Marks
Evaluate the following integrals:$\int\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\text{dx}$
Answer$\int\Big(\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\Big[\frac{\text{x}}{1+\cos\text{x}}+\frac{\sin\text{x}}{1+\cos\text{x}}\Big]\text{dx}$
$=\int\bigg[\frac{\text{x}}{2\cos^{2}\frac{\text{x}}{2}}+\frac{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\bigg]\text{dx}$
$=\frac{1}{2}\int\text{x}\cdot\sec^2\frac{\text{x}}{2}\text{dx}+\int\tan\frac{\text{x}}{2}\text{dx}$
$=\frac{1}{2}\bigg[\text{x}\cdot\frac{\tan\big(\frac{\text{x}}{2}\big)}{\frac{1}{2}}-\int1\times2\tan\big(\frac{\text{x}}{2}\big)\text{dx}\bigg]+\frac{\log\big|\sec\frac{\text{x}}{2}\big|}{\frac{1}{2}}+\text{C}$
$=\text{x}\tan\big(\frac{\text{x}}{2}\big)-\frac{\log\big|\sec\frac{\text{x}}{2}\big|}{\frac{1}{2}}+\log\frac{\big|\sec\frac{\text{x}}{2}\big|}{\frac{1}{2}}+\text{C}$
$=\text{x}\tan\big(\frac{\text{x}}{2}\big)+\text{C}$
View full question & answer→Question 553 Marks
Evaluate:
$\int\frac{\text{e}^{\log\sqrt{\text{x}}}}{\text{x}}\text{dx}$
Answer$\int\frac{\text{e}^{\log\sqrt{\text{x}}}}{\text{x}}\text{dx}=\int\frac{\sqrt{\text{x}}}{\text{x}}\text{dx}$
$=\int\text{x}^\frac{1}{2}\times\text{x}^{-1}\text{dx}$
$=\int\text{x}^{\frac{1}{2}-1}\text{dx}$
$=\int\frac{\text{x}^{\frac{-1}{2}+1}+\text{c}}{\frac{-1}{2}+1}$
$=\frac{\text{x}^\frac{1}{2}}{\frac{1}{2}}$
$=\sqrt{\text{x}}+\text{c}$
View full question & answer→Question 563 Marks
Evaluate the following integrals:
$\int(\text{a}\tan\text{x}+\text{b}\cot \text{x})^2\text{dx}$
Answer$\int(\text{a}\tan\text{x}+\text{b}\cot \text{x})^2\text{dx}$
$=\int(\text{a}^2\tan^2\text{x}+\text{b}^2\cot^2\text{x}+2\text{ab}\tan\text{x}\cot\text{x})\text{dx}$
$=\text{a}^2\int\tan^2\text{x dx}+\text{b}^2\int\cot^2\text{x dx}+2\text{ab}\int\text{dx}$
$=\text{a}^2\int(\sec^2\text{x}-1)\text{dx}+\text{b}^2\int(\text{cosec}^2\text{x}-1)\text{dx}+2\text{ab}\int\text{dx}$
$=\text{a}^2[\tan\text{x}-\text{x}]+\text{b}^2[-\cot\text{x}-\text{x}]+2\text{ab}\text{x}+\text{C}$
$=\text{a}^2\tan\text{x}-\text{b}^2\cot\text{x}-(\text{a}^2+\text{b}^2-2\text{ab})\text{x}+\text{C}$
View full question & answer→Question 573 Marks
Integrate the function in Exercise.
$\sqrt{1+\frac{\text{x}^2}{9}}$
Answer$\int\sqrt{1+\frac{\text{x}^2}{9}}\text{dx}=\int\sqrt{\frac{9+\text{x}^2}{9}}\text{dx}=\int\frac{\sqrt{\text{x}^2+3^2}}{3}\text{dx}=\frac{1}{3}\int\sqrt{\text{x}^2+3^2}\text{dx}$
$=\frac{1}{3}\Bigg[\frac{\text{x}}{2}\sqrt{\text{x}^2+3^2}+\frac{3}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+3^2}\Big|\Bigg]+\text{c}\Bigg[\therefore\ \int\sqrt{\text{x}^2+\text{a}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2+\text{a}^2}+\frac{\text{a}^2}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\Big|\Bigg]+\text{c}$
$=\frac{\text{x}}{6}\sqrt{\text{x}^2+9}+\frac{3}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+9}\Big|+\text{c}$
View full question & answer→Question 583 Marks
Evaluate the following integrals:
$\int\sin^{-1}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)\text{dx}$
Answer$\int\sin^{-1}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)\text{dx}$
$=\int\sin^{-1}(\sin2\text{x} )\text{dx}$ $\Big[\because\ \sin2\text{x}=\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big]$
$=\int2\text{ x dx}$
$=2\int\text{x dx}$
$=\frac{2\text{x}^2}{2}+\text{C}$
$=\text{x}^2+\text{C}$
$\therefore\ \int\sin^{-1}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)\text{dx}=\text{x}^2+\text{C}$
View full question & answer→Question 593 Marks
Integrate the function in Exercise:$\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}$
Answer$\text{I}=\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
$\text{Let x}=\cos\theta\Rightarrow\text{dx}=-\sin\theta\text{d}\theta$
$\text{I}=\int\tan^{-1}\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}(-\sin\theta\text{d}\theta)$
$=-\int^{-1}\sqrt{\frac{2\sin^{2}\frac{\theta}{2}}{2\cos^{2}\frac{\theta}{2}}}\sin\theta\text{d}\theta$
$=-\int\tan^{-1}\tan\frac{\theta}{2}.\sin\theta\text{d}\theta$
$=-\frac{1}{2}\big[\theta.(-\cos\theta)-\int1.(-\cos\theta)\text{d}\theta\big]$
$=-\frac{1}{2}[-\theta\cos\theta+\sin\theta]$
$=+\frac{1}{2}\theta\cos\theta-\frac{1}{2}\sin\theta$
$=\frac{1}{2}\cos^{-1}\text{x}.\text{x}-\frac{1}{2}\sqrt{1-\text{x}^{2}}+\text{C}$
$=\frac{\text{x}}{2}\cos^{-1}\text{x}-\frac{1}{2}\sqrt{1-\text{x}^{2}}+\text{C}$
$=\frac{1}{2}\Big(\text{x}\cos^{-1}\text{x}-\sqrt{1-\text{x}^{2}}\Big)+\text{C}$
View full question & answer→Question 603 Marks
Evaluate the following intregals:
$\int\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
$=\int\sqrt{\frac{(1-\text{x})(1-\text{x})}{(1+\text{x})(1-\text{x})}}\text{dx}$
$=\int\Big(\frac{1-\text{x}}{\sqrt{1-\text{x}^2}}\Big)\text{dx}$
$=\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}}-\int\frac{\text{x dx}}{\sqrt{1-\text{x}^2}}$
Putting $1-\text{x}^2=\text{t}$
$\Rightarrow-2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\text{x dx}=-\frac{\text{dt}}{2}$
Then
$\text{I}=\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}}+\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\sin^{-1}(\text{x})+\frac{1}{2}\times2\sqrt{\text{t}}+\text{C}$
$=\sin^{-1}(\text{x})+\sqrt{1-\text{x}^2}+\text{C}$
View full question & answer→Question 613 Marks
Evaluate the following integrals:
$\int\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\text{dx}$
Answer$\int\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\text{dx}$
$=\int\frac{2\sin^2\text{x}}{2\cos^2\text{x}}\text{dx}$
$=\int\tan^2\text{x dx}$
$=\int(\sec^2\text{x}-1)\text{dx}$
$=\int\sec^2\text{x dx}-\int\text{dx}$
$=\tan\text{x}-\text{x}+\text{C}$
$\therefore\ \int\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\text{dx}=\tan\text{x}-\text{x}+\text{C}$
View full question & answer→Question 623 Marks
Evalute the following integrals:
$\int\frac{\cos\text{x}}{\cos(\text{x}-\text{a})}\text{dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}}{\cos(\text{x}-\text{a})}\text{dx}$
Putting x - a = t
⇒ x = a + t
⇒ dx = dt
$\therefore\text{I}=\int\frac{\cos(\text{a}+\text{t})\text{dt}}{\cos\text{t}}$
$=\frac{\cos\text{a}\cos\text{t}}{\cos\text{t}}-\frac{\sin\text{a}\sin\text{t}}{\cos\text{t}}\text{dt}$
$=\int\big(\cos\text{a}-\sin\text{a}\tan\text{t}\big)\text{dt}$
$=\text{t}\cos\text{a}-\sin\text{a ln}|\sec\text{t}|+\text{C}$
$=(\text{x}-\text{a})\cos\text{a}-\sin\text{a ln}|\sec(\text{a}-\text{a})|+\text{C}\ \big[\text{t}=\text{x}-\text{a}\big]$
View full question & answer→Question 633 Marks
Integrate the rational function in exercise:
$\frac{1}{\text{x}(\text{x}^\text{n}+1)}$
[Hint: multiply numerator and denominator by $x^{n – 1}$ and put $x^n = t$]
Answer$\text{I}=\int\frac{1}{\text{x}(\text{x}^\text{n}+1)}\text{dx}$
Multiplying both numerator and denominator by $nx^{n-1}$,
$\Big[\therefore \ \frac{\text{d}}{\text{dx}}(\text{x}^\text{n}+1)=\text{nx}^{\text{n}-1}\Big]$
$\text{I}=\int\frac{\text{nx}^{\text{n}-1}}{\text{nx}^{\text{n}-1}.\text{x}(\text{x}^\text{n}+1)}\text{dx}=\frac{1}{\text{n}}\int\frac{\text{nx}^{\text{n}-1}}{\text{x}^{\text{n}}(\text{x}^\text{n}+1)}\text{dx}$
Putting $x^n = t$
$\Rightarrow nx^{n-1} = \frac{\text{dx}}{\text{dt}}$
$\Rightarrow nx^{n-1} dx = dt$
$\therefore$ From eq. (i),
$\text{I}=\frac{1}{\text{n}}\int\frac{\text{dt}}{\text{t}(\text{t}+1)}=\frac{1}{\text{n}}\int\frac{1}{\text{t}(\text{t}+1)}\text{dt}=\frac{1}{\text{n}}\int\frac{\text{t}+1-\text{t}}{\text{t}(\text{t}+1)}\text{dt}$
$=\frac{1}{\text{n}}\int\frac{\text{t}+1}{\text{t}(\text{t}+1)}-\int\frac{1}{\text{t}(\text{t}+1)}\text{dt}=\frac{1}{\text{n}}\Bigg[\int\frac{1}{\text{t}}\text{dt}+\int\frac{1}{\text{t}+1}\text{dt}\Bigg]$
$=\frac{1}{\text{n}}\Big[\text{log}|\text{t}|-\text{log}|\text{t}+1|\Big]+\text{c}=\frac{1}{\text{n}}\Big[\text{log}|\text{x}^\text{n}|-\text{log}|\text{x}^\text{n}+1|\Big]+\text{c}$
$=\frac{1}{\text{n}}\text{log}\Bigg|\frac{\text{x}^\text{n}}{\text{x}^\text{n}+1}\Bigg|+\text{c}$
View full question & answer→Question 643 Marks
$\int\frac{1}{\cos^2\text{x}(1-\tan\text{x})^2}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\cos^2\text{x}(1-\tan\text{x})^2}\text{dx}$
$=\int\frac{\sec^2\text{x}}{(1-\tan\text{x})^2}\text{dx}$
$=\int\frac{\sec^2\text{x dx}}{(1-\tan\text{x})^2}$
$-\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\sec^2\text{x dx}=-\text{dt}$
$\therefore\text{I}=\int\frac{-\text{dt}}{\text{t}^2}$
$=-\int\text{t}^{-2}\text{dt}$
$=-\bigg[\frac{\text{t}^{-2+1}}{-2+1}\bigg]+\text{c}$
$=\frac{1}{\text{t}}+\text{c}$
$=\frac{1}{1-\tan\text{x}}+\text{c}$
View full question & answer→Question 653 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^2+6\text{x}+13}\text{dx}$
AnswerWe have $\text{x}^2+6\text{x}+13=\text{x}^2+6\text{x}+3^2-3^2+13=(\text{x}+3)^2+4$
Sol, $\int\frac{\text{dx}}{\text{x}^2+6\text{x}+13}=\int\frac{1}{(\text{x}+3)^2+2^2}\text{dx}$
Let $\text{x}+3=\text{t}$ Then $\text{dx = dt}$
Therefore, $\int\frac{\text{dx}}{\text{x}^2+6\text{x}+13}=\int\frac{\text{dt}}{\text{t}^2+2^2}=\frac{1}{2}\tan^{-1}\frac{\text{t}}{2}+\text{c}$ [by 7.4 (3)]
$=\frac{1}{2}\tan^{-1}\frac{\text{x}+3}{2}+\text{C}$
View full question & answer→Question 663 Marks
Evalute the following integrals:
$\int\frac{\sec\text{x}}{\sec2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sec\text{x}}{\sec2\text{x}}\text{dx},$ then
$\text{I}=\int\frac{\frac{1}{\cos\text{x}}}{\frac{1}{\cos2\text{x}}}\text{dx}$
$=\int\frac{\cos2\text{x}}{\cos\text{x}}\text{dx}$
$=\int\frac{2\cos^2\text{x}-1}{\cos\text{x}}\text{dx}$
$=\int2\cos\text{ x dx}-\int\frac{1}{\cos\text{x}}\text{dx}$
$=2\int\cos\text{dx}-\int\sec\text{x dx}$
$=2\sin\text{x}-\log|\sec\text{x}+\tan\text{x}|+\text{C}$
$\because\text{I}=2\sin\text{x}-\log|\sec\text{x}+\tan\text{x}|+\text{C}$
View full question & answer→Question 673 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\pi}\Big(\sin^2\frac{\text{x}}{2}-\cos^2\frac{\text{x}}{2}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\pi}\Big(\sin^2\frac{\text{x}}{2}-\cos^2\frac{\text{x}}{2}\Big)\text{dx}$
$=-\int_{0}^\limits{\pi}\Big(\cos^2\frac{\text{x}}{2}-\sin^2\frac{\text{x}}{2}\big)\text{dx}$
$=-\int_{0}^\limits{\pi}\cos\text{x dx}$
$\int\cos\text{x dx}=\sin\text{x}=\text{F(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}(\pi)-\text{F}(0)$
$=\sin(\pi)-\sin0$
$=0$
View full question & answer→Question 683 Marks
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{2}}_{0}|\cos2\text{x}|\text{dx}$
Answer$\int^\limits{\frac{\pi}{2}}_{0}|\cos2\text{x}|\text{dx}$
We know that,
$|\cos2\text{x}|=\begin{cases}-\cos2\text{x},&\frac{\pi}{4}\leq\text{x}\leq\frac{\pi}{2}\\\cos2\text{x},&0<\text{x}\leq\frac{\pi}{4}\end{cases}$
$\therefore\ \text{I}=\int^\limits{2}_{-2}|\cos2\text{x}|\text{dx}$
$\Rightarrow\text{I}=\int^\limits{\frac{\pi}{4}}_{0}\cos2\text{x }\text{dx}-\int^\limits{\frac{\pi}{2}}_{\frac{\pi}{4}}\cos2\text{x }\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{4}}_0-\Big[\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_{\frac{\pi}{4}}$
$\Rightarrow\text{I}=\frac{1}{2}-0-0+\frac{1}{2}$
$\Rightarrow\text{I}=1$
View full question & answer→Question 693 Marks
Evaluate the following integrals:$\int\log_{10}\text{x dx}$
AnswerLet $\text{I}=\int\log_{10}\text{x dx}$
$=\int\frac{\log\text{x}}{\log10}\text{dx}$
$=\frac{1}{\log10}\int1\times\log\text{x dx}$
Using integration by parts,
$=\frac{1}{\log10}\Big[\log\text{x}\int\text{dx}-\int\Big(\frac{1}{\text{x}}\int\text{dx}\Big)\text{dx}\Big]$
$=\frac{1}{\log10}\Big[\text{x}\log\text{x}-\int\big(\frac{\text{x}}{\text{x}}\big)\text{dx}\Big]$
$=\frac{1}{\log10}[\text{x}\log\text{x}-\text{x}]$
$\text{I}=\frac{\text{x}}{\log10}(\log\text{x}-1)$
View full question & answer→Question 703 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{1}\frac{1}{\sqrt{1+\text{x}}-\sqrt{\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\frac{1}{\sqrt{1+\text{x}}-\sqrt{\text{x}}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{1}\bigg(\frac{1}{\sqrt{1+\text{x}}-\sqrt{\text{x}}}\times\frac{\sqrt{1+\text{x}}+\sqrt{\text{x}}}{\sqrt{1+\text{x}}+\sqrt{\text{x}}}\bigg)\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{1}\frac{\sqrt{1+\text{x}}-\sqrt{\text{x}}}{1+\text{x}-\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{1}\Big({\sqrt{1+\text{x}}+\sqrt{\text{x}}}\Big)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{2}{3}(1+\text{x})^{\frac{3}{2}}+\frac{2}{3}\text{x}^{\frac{3}{2}}\Big]^1_0$
$\Rightarrow\text{I}=\frac{2}{3}\times2\sqrt{2}+\frac{2}{3}-\frac{2}{3}$
$\Rightarrow\text{I}=\frac{4\sqrt{2}}{3}$
View full question & answer→Question 713 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x }\sin2\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x }\sin2\text{x}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}2\sin^2\text{x }\cos\text{x}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}2(1-\cos^2\text{x})\cos\text{x}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\big(2\cos\text{x}-2\cos^3\text{x}\big)\text{dx}$
$\Rightarrow\text{I}=\bigg[2\sin\text{x}-2\Big(\sin\text{x}-\frac{\sin^3\text{x}}{3}\Big)\bigg]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=\Big[2-2\Big(1-\frac{1}{3}\Big)\Big]-0$
$\Rightarrow\text{I}=\frac{2}{3}$
View full question & answer→Question 723 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{16-6\text{x}-\text{x}^2}}\text{ dx}$
Answer$\int\frac{\text{dx}}{\sqrt{16-6\text{x}-\text{x}^2}}$
$=\int\frac{\text{dx}}{\sqrt{16-(\text{x}^2+6\text{x})}}$
$=\int\frac{\text{dx}}{\sqrt{16-(\text{x}^2+6\text{x}+3^2-3^2)}}$
$=\int\frac{\text{dx}}{\sqrt{16+9-(\text{x}+3)^2}}$
$=\int\frac{\text{dx}}{\sqrt{5^2-(\text{x}+3)^2}}$
$=\sin^{-1}\Big(\frac{\text{x}+3}{5}\Big)+\text{C}$
View full question & answer→Question 733 Marks
Evaluate the following integrals:$\int\frac{\text{x}+7}{3\text{x}^2+25\text{x}+28}\text{ dx}$
Answer$\text{I}=\int\frac{\text{x}+7}{3\text{x}^2+25\text{x}+28}\text{ dx}$ $=\int\frac{\text{x}+7}{3\text{x}^2+21\text{x}+4\text{x}+28}\text{ dx}$ $=\int\frac{\text{x}+7}{3\text{x}(\text{x}+7)+4(\text{x}+7)}\text{ dx}$ $=\int\frac{\text{x}+7}{(3\text{x}+4)(\text{x}+7)}\text{ dx}$ $=\int\frac{1}{(3\text{x}+4)}\text{ dx}$$=\frac{1}{3}\ln|3\text{x}+4|+\text{C}$
View full question & answer→Question 743 Marks
Evaluate the following integrals:
$\int\frac{1}{1-\cos\text{x}}\text{dx}$
Answer$\int\frac{1}{1-\cos\text{x}}\text{dx}$
$=\int\frac{1}{1-\cos\text{x}}\times\frac{1+\cos\text{x}}{1+\cos\text{x}}\times\text{dx}$
$=\int\frac{1+\cos\text{x}}{1-\cos^2\text{x}}\times\text{dx}$
$=\int\frac{1+\cos\text{x}}{\sin^2\text{x}}\times\text{dx}$
$=\int\frac{1}{\sin^2\text{x}}\text{dx}+\int\frac{\cos\text{x}}{\sin^2\text{x}}\text{dx}$
$=\int\text{cosec}^2\text{x dx}+\int\cot\text{x}\times\text{cosec x dx}$
$=-\cot\text{x}-\text{cosec x}+\text{C}$
$\therefore\ \int\frac{1}{1-\cos\text{x}}\text{dx}=-\cot\text{x}-\text{cosec x}+\text{C}$
View full question & answer→Question 753 Marks
Verify the following:
$\int\frac{2\text{x}-1}{2\text{x}+3}\text{dx}=\text{x}-\log\big|(2\text{x}+3)^3\big|+\text{c}$
AnswerLet $\int\frac{2\text{x}-1}{2\text{x}+3}\text{dx}$
$=\int\frac{2\text{x}+3-2}{2\text{x}+3}\text{dx}$
$=\int\Big(\frac{2\text{x}+3}{2\text{x}+3}-\frac{2}{2\text{x}+3}\Big)\text{dx}$
$=\int1\text{dx}-4\int\frac{1}{2\text{x}+3}\text{dx}$
$=\text{x}-\frac{4}{2}\log\big|(2\text{x}+3)\big|+\text{C}$ $\Big[\because\int\frac{1}{\text{ax}+\text{b}}\text{dx}=\frac{1}{\text{a}}\log|\text{ax}+\text{b}|\Big]$
$=\text{x}-2\log|(2\text{x}+3)|+\text{C}$
$=\text{x}-\log|(2\text{x}+3)^2|+\text{C}$ $\big[\because\ \text{a}\log\text{b}=\log\text{b}^\text{a}\big]$
View full question & answer→Question 763 Marks
Evaluate the following integrals:
$\int\Big(\frac{1}{\log\text{x}}-\frac{1}{(\log\text{x})^2}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\Big(\tan^{-1}\text{x}+\frac{1}{1+\text{x}^2}\Big)\text{dx}$
Here, $\text{f(x)}=\tan^{-1}\text{x}$ and $\text{f}'\text{(x)}=\frac{1}{1+\text{x}^2}$
and we know thet,
$\int\text{e}^{\text{ax}}(\text{af}(\text{x})+\text{f}'(\text{x}))\text{dx}=\text{e}^{\text{ax}}\text{f(x)+C}$
$\therefore\int\text{e}^{\text{x}}\Big(\tan^{-1}\text{x}+\frac{1}{1+\text{x}^2}\Big)\text{dx}=\text{e}^{\text{x}}\tan^{-1}\text{x + C}$
Thus,
$\text{I}=\text{e}^{\text{x}}\tan^{-1}\text{x + C}$
View full question & answer→Question 773 Marks
Evaluate the following integrals:
$\int\frac{1}{\sqrt{\tan^{-1}\text{x}}.(1+\text{x}^2)}\text{dx}$
Answer$\int\frac{\text{dx}}{\sqrt{\tan^{-1}\text{x}}(1+\text{x}^2)}$
$\text{Let }\tan^{-1}\text{x}=\text{t}$
$\Rightarrow\frac{1}{1+\text{x}^2}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{1+\text{x}^2}\text{dx}=\text{dt}$
$\text{Now,}\int\frac{\text{dx}}{\sqrt{\tan^{-1}\text{x}}(1+\text{x}^2)}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\int\text{t}^{-\frac{1}{2}}\text{dt}$
$=\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{\tan^{-1}\text{x}}+\text{C}$
View full question & answer→Question 783 Marks
Evaluate the following integrals:
$\int2\text{x}\sec^3\big(\text{x}^2+3\big)\tan\big(\text{x}^2+3\big)\text{dx}$
Answer$\int2\text{x}\sec^3\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big)\text{dx}$ $=\int\sec^2\big(\text{x}^2+3\big).\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big)2\text{x}\text{ dx}$ Let $\sec\big(\text{x}^2+3\big)=\text{t}$ $\Rightarrow\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big).2\text{x}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big).2\text{x}\text{ dx}=\text{dt}$Now, $\int\sec^2\big(\text{x}^2+3\big).\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big)2\text{x}\text{ dx}$
$=\int\text{t}^2\text{dt}$
$=\frac{\text{t}^2}{3}+\text{C}$
$=\frac{\sec^2(\text{x}^2+3)}{3}+\text{C}$
View full question & answer→Question 793 Marks
Evaluate the following integrals:
$\int\frac{1+\sin\text{x}}{\sqrt{\text{x}-\cos\text{x}}}\text{dx}$
Answer$\int\frac{1+\sin\text{x}}{\sqrt{\text{x}-\cos\text{x}}}\text{dx}$
$\text{Let},\text{x}-\cos\text{x}=\text{t}$
$\Rightarrow(1+\sin\text{x})=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(1+\sin\text{x})\text{dx}=\text{dt}$
$\text{Now,}\int\frac{1+\sin\text{x}}{\sqrt{\text{x}-\cos\text{x}}}\text{dx}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\int\text{t}^{-\frac{1}{2}}\text{dt}$
$=\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{\text{x}-\cos\text{x}}+\text{C}$
View full question & answer→Question 803 Marks
Evaluate the following integrals:
$\int\sec^42\text{x}\text{ dx}$
Answer$\int\sec^42\text{x}\text{ dx}$
$=\int\sec^22\text{x}.\sec^22\text{x}\text{ dx}$
$=\int(1+\tan^22\text{x}).\sec^22\text{x}\text{ dx}$
Let $\tan2\text{x}=\text{t}$
$\sec^22\text{x}.2\text{dx}=\text{dt}$
$\sec^22\text{x}.\text{dx}=\frac{\text{dt}}{2}$
Now, $\int(1+\tan^22\text{x}).\sec^22\text{x}\text{ dx}$
$=\frac{1}{2}\int(1+\text{t}^2)\text{dt}$
$=\frac{1}{2}\Big[\text{t}+\frac{\text{t}^3}{3}\Big]+\text{C}$
$=\frac{\text{t}}{2}+\frac{\text{t}^3}{6}+\text{C}$
$=\frac{\tan(2\text{x})}{2}+\frac{\tan^3(2\text{x})}{6}+\text{C}$
View full question & answer→Question 813 Marks
Write a value of $\int\text{e}^{\text{x}}(\sin\text{x}+\cos\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\sin\text{x}+\cos\text{x})\text{dx}$
$\because\ \int\text{e}^{\text{x}}\big(\text{f(x})+\text{f}'(\text{x})\big)\text{dx}=\text{e}^{\text{x}}\text{f(x)}+\text{C}$
Here, $\text{f(x)}=\sin\text{x}$ and $\text{f}'(\text{x})=\cos\text{x}$
$\therefore\ \text{I}=\int\text{e}^{\text{x}}(\sin\text{x}+\cos\text{x})\text{dx}=\text{e}^{\text{x}}\sin\text{x}+\text{C}$
View full question & answer→Question 823 Marks
Evaluate the following integrals:
$\int^\limits4_1\text{f(x)}\text{dx},$ Where $\text{f(x)}=\begin{cases}7\text{x}+3,&\text{if }\ 1\leq\text{x}\leq3\\8\text{x},&\text{if }\ 3\leq\text{x}\leq9\end{cases}$
AnswerWe have,
$\int^\limits4_1\text{f(x)}\text{dx},$ Where $\text{f(x)}=\begin{cases}7\text{x}+3,&\text{if }\ 1\leq\text{x}\leq3\\8\text{x},&\text{if }\ 3\leq\text{x}\leq9\end{cases}$
$\text{I}=\int^\limits4_1\text{f(x)}\text{dx}$
$\Rightarrow\text{I}=\int^\limits{3}_1\text{f(x)}\text{dx}+\int^\limits4_3\text{f(x)}\text{dx}$ [Additive property]
$\Rightarrow\text{I}=\int^\limits{3}_1(7\text{x}+3)\text{dx}+\int^\limits{4}_38\text{x dx}$
$\Rightarrow\text{I}=\Big[\frac{7\text{x}^2}{2}+3\text{x}\big]^3_1+\big[4\text{x}^2\big]^4_2$
$\Rightarrow\text{I}=\frac{63}{2}+9-\frac{7}{2}-3+64-36$
$\Rightarrow\text{I}=\frac{56}{2}+34$
$\Rightarrow\text{I}=62$
View full question & answer→Question 833 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}\Big(\frac{\sin\text{x}\cos\text{x}-1}{\sin^2\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\Big(\frac{\sin\text{x}\cos\text{x}-1}{\sin^2\text{x}}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\big[\cot\text{x}-\text{cosec}^2\text{x}\big]\text{dx}$
Here, $\text{f(x)}=\cot\text{x}$
$\Rightarrow\text{f}'\text{(x)}=-\text{cosec}^2\text{x}$
Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\cot\text{x}=\text{t}$
Diff both sides w.r.t x
$\text{e}^{\text{x}}(\cot\text{x}-\text{cosec}^2\text{x})\text{dx = dt}$
$\therefore\text{I}=\int\text{dt}$
$=\text{t}+\text{C}$
$=\text{e}^{\text{x}}\cot\text{x + C}$
View full question & answer→Question 843 Marks
$\int(\text{e}^{\text{x}}+\frac{1}{\text{e}^{\text{x}}})^2\text{dx}$
Answer$\int(\text{e}^{\text{x}}+\frac{1}{\text{e}^{\text{x}}})^2\text{dx}$
$=\int(\text{e}^{2\text{x}}+\frac{1}{\text{e}^{2\text{x}}}2\text{e}^\text{x}\times\frac{1}{\text{e}^{\text{x}}})\text{dx}$
$=\int(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}+2)\text{dx}$
$=\frac{\text{e}^{2\text{x}}}{2}+\frac{\text{e}^{-2\text{x}}}{-2}+2\text{x}+\text{c}$
$=\frac{\text{e}^{2\text{x}}}{2}-\frac{\text{e}^{-2\text{x}}}{2}+2\text{x}+\text{c}$
View full question & answer→Question 853 Marks
Prove the following Exercise:
$\int^{\frac{\pi}{2}}_{0}2\tan^{3}\text{x dx}=1-\log2$
Answer$\text{Let I}=\int^{\frac{\pi}{4}}\limits_{0}2\tan^{3}\text{x dx}$
$\text{I}=\int^{\frac{\pi}{4}}\limits_{0}\tan^{2}\text{x}\ \tan\text{x dx}=2\int^{\frac{\pi}{4}}\limits_{0}(\sec^{2}\text{x}-1)\tan\text{x dx}$
$=2\int^{\frac{\pi}{4}}\limits_{0}\sec^{2}\text{x}\ \tan\text{x dx}-2\int^{\frac{\pi}{4}}\limits_{0}\tan\text{x dx}$
$=2\int\limits^{\frac{\pi}{4}}_0\sec\text{x}.\sec\text{x}\tan\text{x dx}-2\int\limits^{\frac{\pi}{4}}_0\tan\text{x dx}$
$\text{Now,} \sec\text{x}=\text{t}$
$=2\Big[\frac{\text{t}^2}{2}\Big]^{\sqrt{2}}_1-2\int\limits^{\frac{\pi}{4}}_0\tan\text{x dx}$
$=2\Big[\frac{2}{2}-\frac{1}{2}\Big]-2\int\limits^{\frac{\pi}{4}}_0\tan\text{x dx}$
$=1-2\int\limits^{\frac{\pi}{4}}_0\tan\text{x dx}$
Hence, the given result is Proved
View full question & answer→Question 863 Marks
Evalute the following integrals:
$\int\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}\text{dx}$
Answer$\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}=\frac{2\cos\text{x}-3\sin\text{x}}{2(3\cos\text{x}+2\sin\text{x})}$
Let $3\cos\text{x}+2\sin\text{x}=\text{t}$
$(-3\sin\text{x}+2\cos\text{x})\text{dx}=\text{dt}$
$\int\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}\text{dx}=\int\frac{\text{dt}}{2\text{t}}$
$=\frac{1}{2}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{1}{2}\log|\text{t}|+\text{C}$
$=\frac{1}{2}\log|2\sin\text{x}+3\cos\text{x}|=\text{C}$
View full question & answer→Question 873 Marks
Evaluate the following:
$\int\frac{(1+\cos\text{x})}{\text{x}+\sin\text{x}}\text{dx}$
AnswerConsider that, $\text{I}=\int\frac{(1+\cos\text{x})}{\text{x}+\sin\text{x}}\text{dx}$
Let $\text{x}+\sin\text{x}=\text{t}$
$\Rightarrow\ (1+\cos\text{x})\text{dx}=\text{dt}$
Substituting $\text{x}+\sin\text{x}=\text{t}$ and $(1+\cos\text{x})\text{dx}=\text{dt}$ in I, we get
$\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\log|\text{t}|+\text{C}$ $\Big[\because\int\frac{1}{\text{x}}\text{dx}=\log|\text{x}|+\text{C}\Big]$
$=\log\big|(\text{x}+\sin\text{x})\big|+\text{C}$ $[\because\ \text{t}=\text{x}+\sin\text{x}]$
View full question & answer→Question 883 Marks
Find the integrals of the functions in Exercises:
$\tan^4\text{x}$
Answer$\tan^4\text{x}$
$=\tan^2\text{x}\tan^2\text{x}$
$=\big(\sec^2\text{x}-1\big)\tan^2\text{x}$
$=\sec^2\text{x}\tan^2\text{x}-\tan^2\text{x}$
$=\sec^2\text{x}\tan^2\text{x}-\big(\sec^2\text{x}-1\big)$
$=\sec^2\text{x }\tan^2\text{x}-\sec^2\text{x}+1$
$\therefore\int\tan^4\text{x}\text{ dx}=\int\sec^2\text{x }\tan^2\text{x}\text{ dx}-\int\sec^2\text{x}\text{ dx}+\int1\cdot\text{dx}$
$=\int\sec^2\text{x }\tan^2\text{x}\text{ dx}-\tan\text{x}+\text{x}+\text{C} \ \ \ \ ...\text{(1)}$
Consider $\int\sec^2\text{x }\tan^2\text{x}\text{ dx}$
$\text{Let }\tan\text{x}=\text{t}\Rightarrow\sec^2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\int\sec^2\text{x }\tan^2\text{x }\text{dx}=\int\text{t}^2\text{dt}=\frac{\text{t}^3}{3}=\frac{\tan^3\text{x}}{3}$
From equation(1),we obtain
$\int\tan^4\text{x}\text{ dx}=\frac{1}{3}\tan^3\text{x}-\tan\text{x}+\text{x}+\text{C}$
View full question & answer→Question 893 Marks
Evaluate the definite integral in Exercise:
$\int\limits_\frac{\pi}{2}^{\pi}\text{e}^{\text{x}}\bigg(\frac{1-\sin\text{x}}{1+\cos\text{x}}\bigg)\text{dx}$
Answer$\text{I}=\int^{\pi}\limits_{\frac{\pi}{2}}\text{e}^{\text{x}}\bigg(\frac{1-\sin\text{x}}{1+\cos\text{x}}\bigg)\text{dx}$ $\text{I}=\int^{\pi}\limits_{\frac{\pi}{2}}\text{e}^{\text{x}}\Bigg[\frac{1-2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\sin^{2}\frac{\text{x}}{2}}\Bigg]\text{dx}$ $\text{I}=\int^{\pi}\limits_{\frac{\pi}{2}}\text{e}^{\text{x}}\Bigg[\frac{\text{cosec}^{2}\frac{\text{x}}{2}}{2}-\cot\frac{\text{x}}{2}\Bigg]\text{dx}$ $\text{Let f (x)}=-\cot\frac{\text{x}}{2}$ $\Rightarrow\text{f (x)}=-\bigg(-\frac{1}{2}\text{cosec}^{2}\frac{\text{x}}{2}\bigg)=\frac{1}{2}\text{cosec}^{2}\frac{\text{x}}{2}$ $\therefore\ \text{I}=\int^{\pi}\limits_{\frac{\pi}{2}}\text{e}^{\text{x}}\Big[\text{f(x)}+\text{f}'\text{(x)}\Big]\text{dx}$$=\Big[\text{e}^{\text{x}}.\text{f (x) dx}\Big]_{\frac{\pi}{2}}^{\pi}$
$=-\bigg[\text{e}^\text{x}.\cot\frac{\text{x}}{2}\bigg]^{\pi}_{\frac{\pi}{2}}$ $=-\bigg[\text{e}^{\text{x}}\times\cot\frac{\pi}{2}-\text{e}^{\frac{\pi}{2}}\times\cot\frac{\pi}{4}\bigg]$ $=-\bigg[\text{e}^{\text{x}}\times0-\text{e}^{\frac{\pi}{2}}\times1\bigg]$$=\text{e}^{\frac{\pi}{2}}$
View full question & answer→Question 903 Marks
Evaluate the following integrals:
$\int\frac{5\cos^3\text{x}+6\sin^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}\text{dx}$
Answer$\int\bigg(\frac{5\cos^3\text{x}+6\sin^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}\bigg)\text{dx}$
$=\int\bigg(\frac{5\cos^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}+\frac{6\sin^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}\bigg)\text{dx}$
$=\int\Big(\frac{5}{2}\frac{\cos\text{x}}{\sin^2\text{x}}+3\frac{\sin\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$=\frac{5}{2}\int\Big(\frac{\cos\text{x}}{\sin\text{x}}\times\frac{1}{\sin\text{x}}\Big)\text{dx}+3\int\frac{\sin\text{x}}{\cos\text{x}}\times\frac{1}{\cos\text{x}}\text{dx}$
$=\frac{5}{2}\int(\text{cosec x}\cot\text{x})\text{dx}+3\int\sec\text{x}\tan\text{x dx}$
$=\frac{5}{2}(-\text{cosec x})+3\sec\text{x}+\text{C}$
$=-\frac{5}{2}\text{cosec x}+3\sec\text{x}+\text{C}$
View full question & answer→Question 913 Marks
Evaluate the following definite integrals:
$\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{4}}\text{cosec}\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{4}}\text{cosec}\text{x}\text{ dx}$
$\int\text{cosec}\text{x dx}=\log|\text{cosecx}-\cot\text{x}|=\text{F}(\text{x})$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}\Big(\frac{\pi}{4}\Big)-\text{F}\Big(\frac{\pi}{6}\Big)$
$=\log\Big|\text{cosec}\frac{\pi}{4}-\cot\frac{\pi}{4}\Big|-\log\Big|\text{cosec}\frac{\pi}{6}-\cot\frac{\pi}{6}\Big|$
$=\log\big|\sqrt{2}-1\big|-\log\big|2-\sqrt{3}\big|$
$=\log\bigg(\frac{\sqrt{2}-1}{2-\sqrt{3}}\bigg)$
View full question & answer→Question 923 Marks
Evaluate the following integrals:
$\int\frac{\sqrt{16+(\log\text{x})^2}}{\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sqrt{16+(\log\text{x})^2}}{\text{x}}\text{dx}$
Let $\log\text{x}=\text{t}$
$\Rightarrow\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\sqrt{16+\text{t}^2}\text{dt}$
$=\int\sqrt{4^2+\text{t}^2}\text{dt}$
$=\frac{\text{t}}{2}\sqrt{16+\text{t}^2}+\frac{16}{2}\log\big|\text{t}+\sqrt{16+\text{t}^2}\big|+\text{C}$
$\therefore\ \text{I}=\frac{\log\text{x}}{2}\sqrt{16+(\log\text{x})^2}\\+8\log\Big|\log\text{x}+\sqrt{16+(\log\text{x})^2}\Big|+\text{C}$
View full question & answer→Question 933 Marks
Find the integrals of the functions in Exercises:
$\sin4\text{x}\sin8\text{x}$
AnswerIt is known that, $\sin\text{A}\sin\text{B}=\frac{1}{2}\cos(\text{A}-\text{B})-\cos(\text{A}+\text{B})$ $=\int\bigg\{\frac{1}{2}\cos(4\text{x}-8\text{x})-\cos(4\text{x}+8\text{x})\bigg\}\text{ dx}$ $=\frac{1}{2}\int(\cos(-4\text{x})-\cos12\text{x})\text{dx}$ $=\frac{1}{2}\int\big\{(\cos4\text{x}-\cos12\text{x}\big\}\text{dx}$ $=\frac{1}{2}\bigg[\frac{\sin4\text{x}}{4}-\frac{\sin12\text{x}}{12}\bigg]+\text{C}$
View full question & answer→Question 943 Marks
Evaluate the following integrals:
$\int\frac{5\text{x}^4+12\text{x}^3+7\text{x}^2}{\text{x}^2+\text{x}}\text{dx}$
Answer$\int\frac{5\text{x}^4+12\text{x}^3+7\text{x}^2}{\text{x}^2+\text{x}}\text{dx}$
$=\int\frac{5\text{x}^4+7\text{x}^3+5\text{x}^3+7\text{x}^2}{\text{x}^2+\text{x}}\text{dx}$
$=\int\frac{5\text{x}^3+7\text{x}^2+5\text{x}^2+7\text{x}}{\text{x}+1}\text{dx}$
$=\int\frac{5\text{x}^2(\text{x}+1)+7\text{x}(\text{x}+1)}{\text{x}+1}\text{dx}$
$=\int(5\text{x}^2+7\text{x})\text{dx}$
$=\frac{5\text{x}^3}{3}+\frac{7\text{x}^2}{2}+\text{C}$
View full question & answer→Question 953 Marks
Evalute the following integrals:
$\int\frac{\sin2\text{x}}{\sin\Big(\text{x}-\frac{\pi}{6}\Big)\sin\Big(\text{x}+\frac{\pi}{6}\Big)}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin2\text{x}}{\sin\Big(\text{x}-\frac{\pi}{6}\Big)\sin\Big(\text{x}+\frac{\pi}{6}\Big)}\text{dx}$
$=\int\frac{\sin2\text{x}}{\sin^2\text{x}-\sin^2\frac{\pi}{6}}\text{dx}$
$\big[\because \sin(\text{A}+\text{B})\sin(\text{A}-\text{B})=\sin^2\text{A}-\sin^2\text{B}\big]$
$=\int\frac{\sin2\text{x}}{\sin^2\text{x}-\frac{1}{4}}\text{dx}$
Putting $\sin^2\text{x}-\frac{1}{4}=\text{t}$
$\Rightarrow2\sin\text{x}\cos\text{ x dx}=\text{dt}$
$\Rightarrow\sin2\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\text{ln}|\text{t}|+\text{C}$
$=\text{ln}\big|\sin^2\text{x}-\frac{1}{4}\big|+\text{C}\ \Big[\because\text{t}=\sin^2\text{x}-\frac{1}{4}\Big]$
View full question & answer→Question 963 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2\text{dx}}{\text{x}^6-\text{a}^6}\text{dx}$
Answer$\int\frac{\text{x}^2\text{dx}}{\text{x}^6-\text{a}^6}$
Let $\text{x}^3=\text{t}$
$\Rightarrow3\text{x}^2\text{dx = dt}$
$\Rightarrow\text{x}^2\text{dx}=\frac{\text{dt}}{3}$
Now, $\int\frac{\text{x}^2\text{dx}}{\text{x}^6-\text{a}^6}$
$=\frac{1}{3}\int\frac{\text{dt}}{\text{t}^2-(\text{a}^3)^2}$
$=\frac{1}{3}\times\frac{1}{2\text{a}^3}\log\bigg|\frac{\text{t}-\text{a}^3}{\text{t}+\text{a}^3}\Big|+\text{C}$
$=\frac{1}{6\text{a}^3}\log\Big|\frac{\text{x}^3-\text{a}^3}{\text{x}^3+\text{a}^3}\Big|+\text{C}$
View full question & answer→Question 973 Marks
Write a value of $\int\frac{\log\text{x}^{\text{n}}}{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\log\text{x}^{\text{n}}}{\text{x}}\text{ dx}$
Let $\log\text{x}^{\text{n}}=\text{t}$
$\frac{\text{nx}^{\text{n}-1}}{\text{x}^{\text{n}}}\text{ dx}=\text{dt}$
$\frac{\text{n}}{\text{x}}=\text{ dt}$
$\therefore\ \text{I}=\text{n}\int\text{t}\text{ dt}$
$=\text{n}\Big(\frac{\text{t}^2}{2}\Big)+\text{C}$
Putting the value of t
$\text{I}=\frac{\text{n}(\log\text{x}^{\text{n}})^2}{2}+\text{C}$
View full question & answer→Question 983 Marks
Evalute the following integrals:
$\int\frac{\text{a}}{\text{b}+\text{ce}^\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{a}}{\text{b}+\text{ce}^\text{x}}\text{dx}$
Dividing numerator and denomimator by $e^x$
$\Rightarrow\text{I}=\int\frac{\text{ae}^{-\text{x}}}{\text{be}^{-\text{x}}+\text{c}}\text{dx}$
Putting $e^{-x} = t$
$\Rightarrow-\text{e}^{-\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{-\text{x}}\text{dx}=-\text{dt}$
$\therefore\text{I}=\int\frac{-\text{a}}{\text{bt}+\text{c}}\text{dt}$
$=\frac{-\text{a}}{\text{b}}\text{ ln}|\text{bt}+\text{c}|+\text{C}$
$\Big[\because\int\frac{1}{\text{ax}+\text{b}}\text{dx}=\frac{1}{\text{a}}\text{ ln}|\text{ax}+\text{b}|+\text{C}\Big]$
$=\frac{-\text{a}}{\text{b}}\text{ ln}|\text{be}^{-\text{x}}+\text{c}|+\text{C}\ \big[\because\text{t}=\text{e}^{-\text{x}}+\text{C}\big]$
View full question & answer→Question 993 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}}(\log\text{x})^2\text{dx}$
AnswerLet I $=\int\frac{1}{\text{x}}(\log\text{x})^2\text{dx}\ .....(1)$
Let $\log\text{x}=\text{t}$ then,
$\text{d}(\log\text{x})=\text{dt}$
$\Rightarrow\frac{1}{\text{x}}\text{dx}=\text{dt}$
Putting $\log\text{x}=\text{t}$ and $\frac{1}{\text{x}}\text{dx}=\text{dt}$ in equation (1), we get
$\text{I}=\int\text{t}^2\text{dt}$
$=\frac{\text{t}^3}{3}+\text{C}$
$=\frac{(\log\text{x})^3}{3}+\text{C}$
$\therefore\text{I}=\frac{1}{3}(\log\text{x})^3+\text{C}$
View full question & answer→Question 1003 Marks
Evaluate the following integrals:
$\int\sec^6\text{x }\tan\text{x}\text{ dx}$
Answer$\int\sec^6\text{x }\tan\text{x}\text{ dx}$
$\int\sec^6\text{x}.\sec\text{x}\tan\text{x}\text{ dx}$
Let $\sec\text{x}=\text{t}$
$\sec\text{x}\tan\text{x}\text{ dx}=\text{dt}$
Now, $\int\sec^6\text{x}.\sec\text{x}\tan\text{x}\text{ dx}$
$=\int\text{t}^6\text{dt}$
$=\frac{\text{t}^6}{6}+\text{C}$
$=\frac{\sec^6\text{x}}{6}+\text{C}$
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