Question 12 Marks
Integrate the following with respect to the respective variable: $\frac{t^3}{(t+1)^2}$
View full question & answer→Question 22 Marks
Integrate the following with respect to the respective variable: $\frac{x^7}{x+1}$
View full question & answer→Question 32 Marks
Integrate the following with respect to the respective variable: $(x-2)^2 \sqrt{x}$
View full question & answer→Question 42 Marks
Integrate the following w.r.t. x: $\log \left(x^2+1\right)$
View full question & answer→Question 52 Marks
Evaluate the following : $\int \frac{2 x-7}{\sqrt{3 x-2}} d x$
AnswerExpress $(2 x-7)$ in terms of $(3 x-2)$
$
\begin{aligned}
& 2 x-7=\frac{2}{3}(3 x-2)+\frac{4}{3}-7 \\
& =\frac{2}{3}(3 x-2)-\frac{17}{3} \\
& I=\int\left[\frac{\frac{2}{3}(3 x-2)-\frac{17}{3}}{\sqrt{3 x-2}}\right] \cdot d x \\
& =\int\left[\frac{\frac{2}{3}(3 x-2)}{\sqrt{3 x-2}}-\frac{\frac{17}{3}}{\sqrt{3 x-2}}\right] \cdot d x \\
& =\frac{2}{3} \int \sqrt{3 x-2} \cdot d x-\frac{17}{3} \int \frac{1}{\sqrt{3 x-2}} \cdot d x \\
& =\frac{2}{3} \int(3 x-2)^{\frac{1}{2}} \cdot d x-\frac{17}{3} \int \frac{1}{\sqrt{3 x-2}} \cdot d x \\
& =\frac{2}{3} \cdot \frac{(3 x-2)^{\frac{3}{2}}}{\left(\frac{3}{2}\right)} \cdot \frac{1}{3}-\frac{17}{3} \cdot 2 \cdot(\sqrt{3 x-2}) \cdot \frac{1}{3}+c \\
& =\frac{4}{27} \cdot(3 x-2)^{\frac{3}{2}}-\frac{34}{9} \cdot(3 x-2)^{\frac{1}{2}}+c \\
&
\end{aligned}
$
View full question & answer→Question 62 Marks
Evaluate the following : $\int \frac{1}{\sqrt{3 x+1}-\sqrt{3 x-5}} \cdot d x$
Answer$\int \frac{1}{\sqrt{3 x+1}-\sqrt{3 x-5}} \cdot d x$
$
\begin{aligned}
= & \int\left(\frac{1}{\sqrt{3 x+1}-\sqrt{3 x-5}}\right) \cdot\left(\frac{\sqrt{3 x+1}+\sqrt{3 x-5}}{\sqrt{3 x+1}+\sqrt{3 x-5}}\right) \cdot d x \\
& =\int \frac{\sqrt{3 x+1}+\sqrt{3 x-5}}{3 x+1-3 x+5} \cdot d x \\
& =\int \frac{\sqrt{3 x+1}+\sqrt{3 x-5}}{6} \cdot d x \\
& =\frac{1}{6} \cdot \int\left((3 x+1)^{\frac{1}{2}}+(3 x-5)^{\frac{1}{2}}\right) \cdot d x \\
& =\frac{1}{6} \cdot\left\{\int(3 x+1)^{\frac{1}{2}} \cdot d x+\int(3 x-5)^{\frac{1}{2}} \cdot d x\right\} \\
& =\frac{1}{6} \cdot\left\{\frac{(3 x+1)^{\frac{1}{2}}+1}{\left(\left(\frac{1}{2}+1\right) \cdot 3\right.}+\frac{(3 x-5)^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right) \cdot 3}\right\}+c \\
& =\frac{1}{18} \cdot\left\{\frac{2}{3}(3 x+1)^{\frac{3}{2}}+\frac{2}{3}(3 x-5)^{\frac{3}{2}}\right\}+c \\
& =\frac{1}{27} \cdot\left\{(3 x+1)^{\frac{3}{2}}+(3 x-5)^{\frac{3}{2}}\right\}+c
\end{aligned}
$
View full question & answer→Question 72 Marks
Evaluate : $\int e^x\left[\frac{x+2}{(x+3)^2}\right] \cdot d x$
Answer$
\begin{aligned}
\mathrm{I} & =\int e^x\left[\frac{x+3-1}{(x+3)^2}\right] \cdot d x \\
& =\int e^x\left[\frac{x+3}{(x+3)^2}+\frac{-1}{(x+3)^2}\right] \cdot d x \\
& =\int e^x\left[\frac{1}{x+3}+\frac{-1}{(x+3)^2}\right] \cdot d x \\
& \therefore f(x)=\frac{1}{x+3} \Rightarrow f^{\prime}(x)=\frac{-1}{(x+3)^2} \\
& \therefore \int e^x\left[f(x)+f^{\prime}(x)\right] \cdot d x=e^x \cdot f(x)+c \\
& =e^x \cdot\left(\frac{1}{x+3}\right)+c \\
& =\frac{e^x}{x+3}+c \\
\therefore \quad & \int e^x\left[\frac{x+2}{(x+3)^2}\right] \cdot d x=\frac{e^x}{x+3}+c
\end{aligned}
$
View full question & answer→Question 82 Marks
Evaluate : $\int e^x\left(\frac{2+\sin 2 x}{1+\cos 2 x}\right) \cdot d x$
Answer$
\begin{aligned}
\text { I } & =\int e^x\left(\frac{2+2 \sin x \cdot \cos x}{2 \cdot \cos ^2 x}\right) \cdot d x \\
& =\int e^x\left(\frac{1}{\cos ^2 x}+\frac{\sin x \cdot \cos x}{\cos ^2 x}\right) \cdot d x \\
& =\int e^x\left[\sec ^2 x+\tan x\right] \cdot d x \\
& =\int e^x\left[\tan x+\sec ^2 x\right] \cdot d x \\
& \therefore f(x)=\tan x \Rightarrow f^{\prime}(x)=\sec ^2 x \\
& \therefore \int e^x\left[f(x)+f^{\prime}(x)\right] \cdot d x=e^x \cdot f(x)+c \\
& =e^{x \cdot \tan x+c} \\
\therefore & \int e^x\left(\frac{2+\sin 2 x}{1+\cos 2 x}\right) \cdot d x=e^x \cdot \tan x+c
\end{aligned}
$
View full question & answer→Question 92 Marks
Evaluate : $\int \cos ^{-1} \sqrt{x} \cdot d x$
View full question & answer→Question 102 Marks
Evaluate : $\int \frac{1}{\sqrt{3 x^2-7}} \cdot d x$
Answer$
\begin{aligned}
& \text {I }=\int \frac{1}{\sqrt{3\left(x^2-\frac{7}{3}\right)} \cdot d x} \\
&=\int \frac{1}{\sqrt{3} \cdot \sqrt{x^2-\left(\frac{\sqrt{7}}{\sqrt{3}}\right)^2}} \cdot d x \\
&=\frac{1}{\sqrt{3}} \cdot \int \frac{1}{\sqrt{x^2-\left(\frac{\sqrt{7}}{\sqrt{3}}\right)^2}} \cdot d x \\
& \int \frac{1}{\sqrt{x^2-a^2}} \cdot d x=\log \left|x+\sqrt{x^2-a^2}\right|+c \\
& \mathrm{I}=\frac{1}{\sqrt{3}} \cdot \log \left(x+\sqrt{x^2-\left(\frac{\sqrt{7}}{\sqrt{3}}\right)^2}\right)+c \\
&=\frac{1}{\sqrt{3}} \cdot \log \left(x+\sqrt{x^2-\frac{7}{3}}\right)+c
\end{aligned}
$
View full question & answer→Question 112 Marks
Evaluate : $\int \frac{1}{a^2-b^2 x^2} \cdot d x$
Answer\begin{aligned}
& \mathrm{I}=\int \frac{1}{b^2\left(\frac{a^2}{b^2}-x^2\right)} \cdot d x \\
& =\frac{1}{b^2} \cdot \int \frac{1}{\left(\frac{a}{b}\right)^2-x^2} \cdot d x \\
& \because \quad \int \frac{1}{a^2-x^2} \cdot d x=\frac{1}{2 a} \log \left(\frac{a+x}{a-x}\right)+c \\
& \mathrm{I}=\frac{1}{b^2} \cdot \frac{1}{2\left(\frac{a}{b}\right)} \cdot \log \left(\frac{\frac{a}{b}+x}{\frac{a}{b}-x}\right)+c \\
& =\frac{1}{b^2} \cdot \frac{1}{2\left(\frac{a}{b}\right)} \cdot \log \left(\frac{\frac{a}{b}+x}{\frac{a}{b}-x}\right)+c \\
& =\frac{1}{2 a b} \cdot \log \left(\frac{a+b x}{a-b x}\right)+c \\
\end{aligned}
View full question & answer→Question 122 Marks
Evaluate : $\int \frac{1}{4 x^2+11} \cdot d x$
Answer$
\begin{aligned}
& \text { Solution : I }=\int \frac{1}{4\left(x^2+\frac{11}{4}\right)} \cdot d x \\
& =\frac{1}{4} \cdot \int \frac{1}{x^2+\left(\frac{\sqrt{11}}{2}\right)^2} \cdot d x \\
& \because \quad \int \frac{1}{x^2+a^2} \cdot d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \\
& \mathrm{I}=\frac{1}{4} \cdot\left[\frac{1}{\left(\frac{\sqrt{11}}{2}\right)}\right] \cdot \tan ^{-1}\left[\frac{x}{\left(\frac{\sqrt{11}}{2}\right)}\right]+c \\
& =\frac{1}{2 \sqrt{11}} \tan ^{-1}\left(\frac{2 x}{\sqrt{11}}\right)+c \\
\end{aligned}
$
View full question & answer→Question 132 Marks
Prove The Following : $\int \frac{1}{x \sqrt{x^2-a^2}} \cdot d x=\frac{1}{a} \sec ^{-1}\left(\frac{x}{a}\right)+c$
AnswerLet $\mathrm{I}=\int \frac{1}{x \sqrt{x^2-a^2}} \cdot d x$ put $x=a \sec \theta \Rightarrow \quad \theta=\sec ^{-1}\left(\frac{x}{a}\right)$
$
\begin{aligned}
\therefore \quad d x & =a \cdot \sec \theta \cdot \tan \theta \cdot d \theta \\
\mathrm{I} & =\int \frac{1}{a \sec \theta \sqrt{\ldots-a^2}} \cdot \ldots \ldots \\
& =\int \frac{\tan \theta}{\sqrt{a^2(\ldots \ldots)}} \cdot d \theta \\
& =\frac{1}{a} \int 1 \cdot d \theta \\
& =\frac{1}{a} \cdot \theta+c \\
& =\frac{1}{a} \cdot \sec ^{-1}\left(\frac{x}{a}\right)+c \\
\therefore \quad & \int \frac{1}{x \sqrt{x^2-a^2}} \cdot d x=\frac{1}{a} \sec ^{-1}\left(\frac{x}{a}\right)+c \\
\text { e.g. } \quad & \int \frac{1}{x \sqrt{x^2-64}} \cdot d x=\frac{1}{8} \sec ^{-1}\left(\frac{x}{8}\right)+c
\end{aligned}
$
View full question & answer→Question 142 Marks
Prove The Following : $\int \frac{1}{\sqrt{a^2-x^2}} \cdot d x=\sin ^{-1}\left(\frac{x}{a}\right)+c$
AnswerLet $I=\int \frac{1}{\sqrt{a^2-x^2}} \cdot d x$ put
$
\begin{gathered}
x=a \sin \theta \Rightarrow \quad \sin \theta=\frac{x}{a} \\
\therefore \quad \theta=\sin ^{-1}\left(\frac{x}{a}\right) \\
d x=a \cdot \cos \theta d \theta \\
\mathrm{I} \quad=\int \frac{1}{\sqrt{a^2-a^2 \sin ^2 \theta}} \cdot a \cdot \cos \theta d \theta \\
\mathrm{I} \quad=\int \frac{a \cdot \cos \theta}{a \sqrt{1-a^2 \sin ^2 \theta}} \cdot d \theta
\end{gathered}
$
$\begin{aligned} &=\int \frac{\cos \theta}{\cos \theta} \cdot d \theta \\ &=\int 1 \cdot d \theta \\ &=\theta+c \\ & \therefore \quad \int \frac{1}{\sqrt{a^2-x^2}} \cdot d x=\sin ^{-1}\left(\frac{x}{a}\right)+c \\ & \text { e.g. } \int \frac{1}{\sqrt{81-x^2}} \cdot d x=\sin ^{-1}\left(\frac{x}{9}\right)+c\end{aligned}$
View full question & answer→Question 152 Marks
Prove The Following : $\int \frac{1}{a^2-x^2} \cdot d x=\frac{1}{2 a} \log \left(\frac{a+x}{a-x}\right)+c$
AnswerConsider,
$
\begin{aligned}
& \mathrm{I}=\int \frac{1}{a^2-x^2} \cdot d x \\
&=\int \frac{1}{(\ldots)(\ldots)} \cdot d x \\
&=\int \frac{1}{2 a} \cdot\left[\frac{1}{\ldots}-\frac{1}{\ldots}\right] \cdot d x \\
&=\frac{1}{2 a} \cdot \int\left[\frac{1}{\ldots}-\frac{1}{a+x}\right] \cdot d x \\
&=\frac{1}{2 a} \cdot[\log (a+x)-\log (a-x)]+c \\
&=\frac{1}{2 a} \cdot \log \left(\frac{\ldots}{\ldots}\right)+c \\
& \therefore \quad \int \frac{1}{a^2-x^2} \cdot d x=\frac{1}{2 a} \log \left(\frac{a+x}{a-x}\right)+c \\
& \text { e.g. } \quad \int \frac{1}{16-x^2} \cdot d x=\frac{1}{2(4)} \log \left(\frac{4+x}{4-x}\right)+c
\end{aligned}
$
View full question & answer→Question 162 Marks
Prove The Following : $\int \frac{1}{x^2-a^2} \cdot d x=\frac{1}{2 a} \log \left(\frac{x-a}{x+a}\right)+c$
Answer$
\begin{aligned}
& \text { Let } \quad I=\int \frac{1}{x^2-a^2} \cdot d x \\
& =\int \frac{1}{(x+a)(x-a)} \cdot d x \\
& =\int \frac{1}{2 a} \cdot\left[\frac{1}{x-a}-\frac{1}{x+a}\right] \cdot d x \\
& =\frac{1}{2 a} \cdot \int\left[\frac{1}{x-a}-\frac{1}{x+a}\right] \cdot d x \\
& =\frac{1}{2 a} \cdot[\log (x-a)-\log (x+a)]+c \\
& =\frac{1}{2 a} \cdot \log \left(\frac{x-a}{x+a}\right)+c \\
& \therefore \quad \int \frac{1}{x^2-a^2} \cdot d x=\frac{1}{2 a} \log \left(\frac{x-a}{x+a}\right)+c \\
& \text { e.g. } \int \frac{1}{x^2-9} \cdot d x=\frac{1}{2(3)} \log \left(\frac{x-3}{x+3}\right)+c \\
&
\end{aligned}
$
View full question & answer→Question 172 Marks
Evaluate the following functions : $\int \frac{\sin (x+a)}{\cos (x-b)} \cdot d x$
Answer$
\begin{aligned}
& =\int \frac{\sin [(x-b)+(a+b)]}{\cos (x-b)} \cdot d x \\
& =\int \frac{\sin (x-b) \cdot \cos (a+b)+\cos (x-b) \cdot \sin (a+b)}{\cos (x-b)} \cdot d x \\
& =\int\left[\frac{\sin (x-b) \cdot \cos (a+b)}{\cos (x-b)}+\frac{\cos (x-b) \cdot \sin (a+b)}{\cos (x-b)}\right] \\
& =\int[\cos (a+b) \cdot \tan (x-b)+\sin (a+b)] \cdot d x \\
& =\cos (a+b) \cdot \log (\sec (x-b))+x \cdot \sin (a+b)+c
\end{aligned}
$
View full question & answer→Question 182 Marks
Evaluate the following functions : $\int(3 x+2) \sqrt{x-4} \cdot d x$
Answer$
\begin{aligned}
& \text { put } \quad x-4=t \\
& \therefore \quad x=4+t
\end{aligned}
$
Differentiate
$
\begin{aligned}
& 1 \cdot d x=1 \cdot d t \\
= & \int[3(4+t)+2] \cdot \sqrt{t} \cdot d t \\
= & \int(14+3 t) \cdot t^{\frac{1}{2}} \cdot d t \\
= & \int\left(14 t^{\frac{1}{2}}+3 t^{\frac{3}{2}}\right) \cdot d t \\
= & 14 \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+3 \frac{t^{\frac{5}{2}}}{\frac{5}{2}} \cdot d x \\
= & \frac{28}{3}(x-4)^{\frac{3}{2}}+\frac{6}{5}(x-4)^{\frac{5}{2}}+c
\end{aligned}
$
View full question & answer→Question 192 Marks
Evaluate the following functions : $\int \frac{1}{3 x+7 x^{-n}} \cdot d x$
Answer$
\begin{aligned}
& \text { : Consider } \int \frac{1}{3 x+7 x^{-n}} \cdot d x \\
& \quad=\int \frac{1}{3 x+\frac{7}{x^n}} \cdot d x=\int \frac{1}{\frac{3 x^{n+1}+7}{x^n}} \cdot d x \\
& \quad=\int \frac{x^n}{3 x^{n+1}+7} \cdot d x \\
& \quad \text { put } 3 x^{n+1}+7=t \\
& \quad \text { Differentiate } w \cdot r \cdot t \cdot x \\
& \quad 3(n+1) x^n \cdot d x=d t \\
& \therefore \quad x^n \cdot d x=\frac{1}{3(n+1)} d t \\
& =\int \frac{1}{3(n+1)} \cdot d t \\
& =\frac{1}{3(n+1)} \cdot \log (t)+c \\
& =\frac{1}{3(n+1)} \cdot \log \left(3 x^{n+1}+7\right)+c
\end{aligned}
$
View full question & answer→Question 202 Marks
Evaluate the following : $\int \tan ^{-1} \sqrt{\frac{1-\sin x}{1+\sin x}} \cdot d x$
Answer$
\begin{aligned}
I & =\int \tan ^{-1} \sqrt{\frac{1-\cos \left(\frac{\pi}{2}-x\right)}{1+\cos \left(\frac{\pi}{2}-x\right)}} \cdot d x \\
& =\int \tan ^{-1} \sqrt{\frac{2 \sin ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \cos ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}} \cdot d x \\
& =\int \tan ^{-1} \sqrt{\tan ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)} \cdot d x \\
& =\int \tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right] \cdot d x \\
& =\int\left(\frac{\pi}{4}-\frac{x}{2}\right) \cdot d x \\
& =\frac{\pi}{4} x-\frac{1}{2} \cdot \frac{x^2}{2}+c \\
& =\frac{\pi}{4} x-\frac{x^2}{4}+c
\end{aligned}
$
View full question & answer→Question 212 Marks
Evaluate the following : $\int \frac{\cos x-\cos 2 x}{1-\cos x} \cdot d x$
Answer$
\begin{aligned}
& \int \frac{\cos x-\cos 2 x}{1-\cos x} \cdot d x \\
& =\int \frac{\cos x-(\ldots \ldots \ldots)}{1-\cos x} \cdot d x \\
& =\int \frac{\cos x-\ldots \ldots \ldots}{1-\cos x} \cdot d x \\
& =\int \frac{\cos x(1-\cos x)+\ldots \ldots \ldots}{1-\cos x} \cdot d x \\
& =\int\left[\cos x+\frac{\ldots \ldots \ldots}{1-\cos x}\right] \cdot d x \\
& =\int[\cos x+(1+\cos x)] \cdot d x \\
& =\int(1+2 \cos x) \cdot d x \\
& =x+2 \sin x+c
\end{aligned}
$
View full question & answer→Question 222 Marks
Evaluate the following : $\int\left(\frac{\cos x}{1-\cos x}\right) \cdot d x$
Answer$
\begin{aligned}
I & =\int\left(\frac{\cos x}{1-\cos x}\right)\left(\frac{1+\cos x}{1+\cos x}\right) \cdot d x \\
& =\int \frac{\cos x(1+\cos x)}{1-\cos ^2 x} \cdot d x \\
& =\int\left(\frac{\left.\cos x+\cos ^2 x\right)}{\sin ^2 x}\right) \cdot d x \\
& =\int\left(\frac{\cos x}{\sin ^2 x}+\frac{\cos ^2 x}{\sin ^2 x}\right) \cdot d x \\
& =\int\left(\operatorname{cosec} x \cdot \cot x+\cot ^2 x\right) \cdot d x \\
& =\int\left(\operatorname{cosec} x \cdot \cot x+\operatorname{cosec}^2 x-1\right) \cdot d x \\
& =(-\operatorname{cosec} x)+(-\cot x)-x+c \\
& =-\operatorname{cosec} x-\cot x-x+c
\end{aligned}
$
View full question & answer→Question 232 Marks
Evaluate the following : $\int \frac{\sin ^3 x-\cos ^3 x}{\sin ^2 x \cdot \cos ^2 x} \cdot d x$
Answer$\mathrm{I}=\int\left(\frac{\sin ^3 x}{\sin ^2 x \cdot \cos ^2 x}-\frac{\cos ^3 x}{\sin ^2 x \cdot \cos ^2 x}\right) \cdot d x$
$
=\int\left(\frac{\sin x}{\cos ^2 x}-\frac{\cos x}{\sin ^2 x}\right) \cdot d x
$
$\begin{aligned} & =\int\left(\frac{1}{\cos x} \cdot \frac{\sin x}{\cos x}-\frac{1}{\sin x} \cdot \frac{\cos x}{\sin x}\right) \cdot d x \\ & =\int(\sec x \cdot \tan x-\operatorname{cosec} x \cdot \cot x) \cdot d x \\ & =\sec x-(-\operatorname{cosec} x)+c \\ I & =\sec x+\operatorname{cosec} x+c\end{aligned}$
View full question & answer→Question 242 Marks
Evaluate the following : $\int \sin 5 x \cdot \cos 7 x \cdot d x$
AnswerWe know that
$
\begin{aligned}
& 2 \sin A \cdot \cos B=\sin (A+B)+\sin (A-B) \\
I & =\frac{1}{2} \int 2 \sin 5 x \cdot \cos 7 x \cdot d x \\
= & \frac{1}{2} \int[\sin (5 x+7 x)+\sin (5 x-7 x)] \cdot d x \\
= & \frac{1}{2} \int[\sin (12 x)+\sin (-2 x)] \cdot d x \\
= & \frac{1}{2} \int(\sin 12 x-\sin 2 x) \cdot d x \\
= & \frac{1}{2} \cdot\left[-\cos 12 x \cdot \frac{1}{12}+\cos 2 x \cdot \frac{1}{2}\right]+c \\
I= & -\frac{1}{24} \cos 12 x+\frac{1}{4} \cos 2 x+c
\end{aligned}
$
View full question & answer→Question 252 Marks
Evaluate the following : $\int \sin ^4 x \cdot d x$
Answer$
\begin{aligned}
I & =\int\left(\sin ^2 x\right)^2 \cdot d x \\
& =\int\left(\frac{1}{2}(1-\cos 2 x)\right)^2 \cdot d x \\
& =\frac{1}{4} \cdot \int\left(1-2 \cos 2 x+\cos ^2 2 x\right) \cdot d x \\
& =\frac{1}{4} \cdot \int\left[1-2 \cos 2 x+\frac{1}{2}(1+\cos 4 x)\right] \cdot d x \\
& =\frac{1}{4} \cdot \int\left(1-2 \cos 2 x+\frac{1}{2}+\frac{1}{2} \cos 4 x\right) \cdot d x \\
& =\frac{1}{4} \cdot \int\left(\frac{3}{2}-2 \cos 2 x+\frac{1}{2} \cos 4 x\right) \cdot d x \\
& =\frac{1}{4} \cdot\left[\frac{3}{2} x-2 \sin 2 x \cdot \frac{1}{2}+\frac{1}{2} \sin 4 x \cdot \frac{1}{4}\right]+c \\
& =\frac{1}{4} \cdot\left[\frac{3}{2} x-\sin 2 x+\frac{1}{8} \sin 4 x\right]+c
\end{aligned}
$
View full question & answer→Question 262 Marks
Evaluate the following : $\int \sqrt{1+\sin 3 x} \cdot d x$
Answer$
\begin{aligned}
I & =\int \sqrt{\cos ^2 \frac{3 x}{2}+\sin ^2 \frac{3 x}{2}+2 \sin \frac{3 x}{2} \cdot \cos \frac{3 x}{2}} \cdot d x \\
& =\int \sqrt{\left(\cos \frac{3 x}{2}+\sin \frac{3 x}{2}\right)^2 \cdot d x} \\
& =\int\left(\cos \frac{3 x}{2}+\sin \frac{3 x}{2}\right) \cdot d x \\
& =\sin \frac{3 x}{2} \cdot \frac{1}{\frac{3}{2}}-\cos \frac{3 x}{2} \cdot \frac{1}{\frac{3}{2}}+c \\
& =\frac{2}{3}\left(\sin \frac{3 x}{2}-\cos \frac{3 x}{2}\right)+c
\end{aligned}
$
View full question & answer→Question 272 Marks
Integrate the following w. r. t. x:$\frac{2 x}{\left(2+x^2\right)\left(3+x^2\right)}$
View full question & answer→Question 282 Marks
Integrate the following w. r. t. x:$\frac{1}{x\left(x^5+1\right)}$
View full question & answer→Question 292 Marks
Integrate the following functions w.r.t. x: $\sqrt{2 x^2+3 x+4}$
View full question & answer→Question 302 Marks
Integrate the following functions w.r.t. x: $\sqrt{4^x\left(4^x+4\right)}$
View full question & answer→Question 312 Marks
Integrate the following functions w.r.t. x: $\sqrt{(x-3)(7-x)}$
View full question & answer→Question 322 Marks
Integrate the following functions w.r.t. x:
cosec (log x)[1 – cot(log x)]
Question 332 Marks
Integrate the following functions w.r.t. x: $\log (1+x)^{(1+x)}$
View full question & answer→Question 342 Marks
Integrate the following functions w.r.t. x: $\frac{e^x}{x} \cdot\left[x(\log x)^2+2 \log x\right]$
View full question & answer→Question 352 Marks
Integrate the following functions w.r.t. x: $\left[\frac{x}{(x+1)^2}\right] e^x$
View full question & answer→Question 362 Marks
Integrate the following functions w.r.t. x: $\left(\frac{1+\sin x}{1+\cos x}\right) e^x$
View full question & answer→Question 372 Marks
Evaluate the following: $\int \cos (\sqrt[3]{x}) d x$
View full question & answer→Question 382 Marks
Evaluate the following: ∫sin θ . log(cos θ) dθ
Question 392 Marks
Evaluate the following: $\int \cos \sqrt{x} d x$
View full question & answer→Question 402 Marks
Evaluate the following: $\int \frac{t \cdot \sin ^{-1} t}{\sqrt{1-t^2}} d t$
View full question & answer→Question 412 Marks
Evaluate the following: $\int \frac{\log (\log x)}{x} d x$
View full question & answer→Question 422 Marks
Evaluate the following: $\int x^3 \log x d x$
View full question & answer→Question 432 Marks
Evaluate the following: $\int x \cdot \sin ^2 x d x$
View full question & answer→Question 442 Marks
Evaluate the following: $\int \sec ^3 x d x$
View full question & answer→Question 452 Marks
Evaluate the following: $\int x^2 \log x d x$
View full question & answer→Question 462 Marks
Evaluate : $\int \frac{3 \cos x}{4 \sin ^2 x+4 \sin x-1} d x$
View full question & answer→Question 472 Marks
Evaluate the following : $\int \frac{1}{4+3 \cos ^2 x} \cdot d x$
View full question & answer→Question 482 Marks
Evaluate the following : $\int \frac{1}{4 x^2-20 x+17} \cdot d x$
View full question & answer→Question 492 Marks
Evaluate the following : $\int \frac{1}{1+x-x^2} \cdot d x$
View full question & answer→Question 502 Marks
Evaluate the following : $\int \frac{1}{\sqrt{3 x^2+8}} \cdot d x$
View full question & answer→Question 512 Marks
Evaluate the following : $\int \frac{1}{7+2 x^2} \cdot d x$
View full question & answer→Question 522 Marks
Evaluate the following : $\int \frac{1}{25-9 x^2} \cdot d x$
View full question & answer→Question 532 Marks
Evaluate the following : $\int \frac{1}{4 x^2-3} \cdot d x$
View full question & answer→Question 542 Marks
Integrate the following functions w. r. t. x$\int \frac{1}{\cos x-\sqrt{3} \sin x} \cdot d x$
View full question & answer→Question 552 Marks
Integrate the following functions w. r. t. x$\int \frac{1}{\cos x-\sin x} \cdot d x$
View full question & answer→Question 562 Marks
Integrate the following functions w. r. t. x$\int \frac{1}{3-2 \cos 2 x} \cdot d x$
View full question & answer→Question 572 Marks
Integrate the following functions w. r. t. x$\int \frac{1}{3+2 \sin x-\cos x} \cdot d x$
View full question & answer→Question 582 Marks
Integrate the following functions w. r. t. x$\int \frac{1}{2+\cos x-\sin x} \cdot d x$
View full question & answer→Question 592 Marks
Integrate the following functions w. r. t. x$\int \frac{1}{4-5 \cos x} \cdot d x$
View full question & answer→Question 602 Marks
Integrate the following functions w.r.t. x: $\frac{1}{x \cdot \log x \cdot \log (\log x)}$
AnswerLet $I=\int \frac{1}{x \cdot \log x \cdot \log (\log x)} d x$$=\int \frac{1}{\log (\log x)} \cdot \frac{1}{x \cdot \log x} d x$
Put $\log (\log x)=t \quad \therefore \frac{1}{\log x} \cdot \frac{1}{x} d x=d t$
$\therefore \frac{1}{x \cdot \log x} d x=d t$
$\begin{aligned} \therefore I & =\int \frac{1}{t} d t=\log |t|+c \\ & =\log |\log (\log x)|+c .\end{aligned}$
View full question & answer→Question 612 Marks
Integrate the following functions w.r.t. x: $\frac{x^2}{\sqrt{9-x^6}}$
AnswerLet $I=\int \frac{x^2}{\sqrt{9-x^6}} d x$Put $x^3=t \quad \therefore 3 x^2 d x=d t \quad \therefore x^2 d x=\frac{1}{3} d t$
$\therefore I=\int \frac{1}{\sqrt{9-t^2}} \cdot \frac{d t}{3}$
$\begin{aligned} & =\frac{1}{3} \int \frac{d t}{\sqrt{3^2-t^2}} \\ = & \frac{1}{3} \sin ^{-1}\left(\frac{t}{3}\right)+c \\ = & \frac{1}{3} \sin ^{-1}\left(\frac{x^3}{3}\right)+c .\end{aligned}$
View full question & answer→Question 622 Marks
Integrate the following functions w.r.t. x:
$(5-3 x)(2-3 x)^{-\frac{1}{2}}$
AnswerLet $I=\int(5-3 x)(2-3 x)^{-\frac{1}{2}} d x$Put $2-3 x=t$
$\therefore-3 d x=d t$
$\therefore d x=\frac{-d t}{3}$
Also, $x=\frac{2-t}{3}$
$ \therefore I =\int\left[5-3\left(\frac{2-t}{3}\right)\right] t^{-\frac{1}{2}} \cdot\left(\frac{-d t}{3}\right)$
$=-\frac{1}{3} \int(5-2+t) t^{-\frac{1}{2}} d t$
$ =-\frac{1}{3} \int(3+t) t^{-\frac{1}{2}} d t$
$ =-\frac{1}{3} \int\left(3 t^{-\frac{1}{2}}+t^{\frac{1}{2}}\right) d t$
$=-\frac{3}{3} \int t^{-\frac{1}{2}} d t-\frac{1}{3} \int t^{\frac{1}{2}} d t$
$=-\frac{t^{\frac{1}{2}}}{(1 / 2)}-\frac{1}{3} \cdot \frac{t^{\frac{3}{2}}}{(3 / 2)}+c$
$=-2 \sqrt{2-3 x}-\frac{2}{9}(2-3 x)^{\frac{3}{2}}+c .$
View full question & answer→Question 632 Marks
Integrate the following functions w.r.t. x:
$x^5 \sqrt{a^2+x^2}$
AnswerLet $I=\int x^5 \sqrt{a^2+x^2} d x$Put, $a^2+x^2=t$
$\therefore 2 x d x=d t \quad \therefore x d x=\frac{1}{2} d t$
Also, $x^2=t-a^2$
$I =\int x^2 \cdot x^2 \sqrt{a^2+x^2} x d x$
$=\frac{1}{2} \int\left(t-a^2\right)^2 \sqrt{t} d t$
$=\frac{1}{2} \int\left(t^2-2 a^2 t+a^4\right) \sqrt{t} d t$
$=\frac{1}{2} \int\left(t^{\frac{5}{2}}-2 a^2 t^{\frac{3}{2}}+a^4 t^{\frac{1}{2}}\right) d t$
$=\frac{1}{2} \int t^{\frac{5}{2}} d t-a^2 \int t^{\frac{3}{2}} d t+\frac{a^4}{2} \int t^{\frac{1}{2}} d t$
$=\frac{1}{2} \cdot \frac{t^{\frac{7}{2}}}{\left(\frac{7}{2}\right)}-a^2 \cdot \frac{t^{\frac{5}{2}}}{\left(\frac{5}{2}\right)}+\frac{a^4}{2} \cdot \frac{t^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}+c$
$=\frac{1}{7}\left(a^2+x^2\right)^{\frac{7}{2}}-\frac{2 a^2}{5}\left(a^2+x^2\right)^{\frac{5}{2}}+\frac{a^4}{3}\left(a^2+x^2\right)^{\frac{3}{2}}+c .$
View full question & answer→Question 642 Marks
Integrate the following functions w.r.t. x:
$(2 x+1) \sqrt{x+2}$
AnswerLet $I=\int(2 x+1) \sqrt{x+2} d x$Put $x+2=t \quad \therefore d x=d t$
Also, $x=t-2$
$ \therefore 2 x+1=2(t-2)+1=2 t-3$
$\therefore I=\int(2 t-3) \sqrt{t} d t$
$=\int\left(2 t^{\frac{3}{2}}-3 t^{\frac{1}{2}}\right) d t$
$=2 \int t^{\frac{3}{2}} d t-3 \int t^{\frac{1}{2}} d t$
$=2 \cdot \frac{t^{\frac{5}{2}}}{\left(\frac{5}{2}\right)}-3 \cdot \frac{t^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}+c$
$=\frac{4}{5}(x+2)^{\frac{5}{2}}-2(x+2)^{\frac{3}{2}}+c$.
View full question & answer→Question 652 Marks
Integrate the following functions w.r.t. x:
$\frac{1}{\sqrt{x}+\sqrt{x^3}}$
AnswerLet $I=\int \frac{1}{\sqrt{x}+\sqrt{x^3}} d x$$=\int \frac{1}{x^{\frac{1}{2}}+x^{\frac{3}{2}}} d x$
Put $x=t^2 \quad \therefore d x=2 t d t$
Also $x^{\frac{1}{2}}=\left(t^2\right)^{\frac{1}{2}}=t$ and $x^{\frac{3}{2}}=\left(t^2\right)^{\frac{3}{2}}=t^3$
$\therefore I =\int \frac{2 t d t}{t+t^3}$
$ =2 \int \frac{t d t}{t\left(1+t^2\right)}$
$=2 \int \frac{1}{1+t^2} d t$
$=2 \tan ^{-1} t+c$
$=2 \tan ^{-1}(\sqrt{x})+c .$
View full question & answer→Question 662 Marks
Integrate the following functions w.r.t. x:
$\frac{(x-1)^2}{\left(x^2+1\right)^2}$
AnswerLet $I=\int \frac{(x-1)^2}{\left(x^2+1\right)^2} d x$$=\int \frac{x^2-2 x+1}{\left(x^2+1\right)^2} d x$
$=\int \frac{\left(x^2+1\right)-2 x}{\left(x^2+1\right)^2} d x$
$=\int\left[\frac{x^2+1}{\left(x^2+1\right)^2}-\frac{2 x}{\left(x^2+1\right)^2}\right] d x$
$=\int \frac{1}{x^2+1} d x-\int \frac{2 x}{\left(x^2+1\right)^2} d x$
$=I_1-I_2$
In $I_2$, Put $x^2+1=t \quad \therefore 2 x d x=d t$
$ \therefore I =\int \frac{1}{x^2+1} d x-\int t^{-2} d t$
$ =\tan ^{-1} x-\frac{t^{-1}}{(-1)}+c$
$ =\tan ^{-1} x+\frac{1}{x^2+1}+c .$
View full question & answer→Question 672 Marks
Integrate the following functions w.r.t. x: $\frac{\sqrt{\tan x}}{\sin x \cdot \cos x}$
AnswerLet $I =\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x$Dividing numerator and denominator by $\cos ^2 x$, we get
$\begin{aligned} I & =\int \frac{\left(\frac{\sqrt{\tan x}}{\cos ^2 x}\right)}{\left(\frac{\sin x}{\cos x}\right)} d x \\ & =\int \frac{\sqrt{\tan x} \cdot \sec ^2 x}{\tan x} d x\end{aligned}$
$=\int \frac{\sec ^2 x}{\sqrt{\tan x}} d x$
Put $\tan x=t \quad \therefore \sec ^2 x d x=d t$
$\begin{aligned} \therefore I & =\int \frac{1}{\sqrt{t}} d t=\int t^{-\frac{1}{2}} d t \\ & =\frac{t^{\frac{1}{2}}}{1 / 2}+c=2 \sqrt{t}+c \\ & =2 \sqrt{\tan x}+c .\end{aligned}$
View full question & answer→Question 682 Marks
Integrate the following functions w.r.t. x:
$e^{3 \log x} \cdot\left(x^4+1\right)^{-1}$
AnswerLet $I=e^{3 \log x}\left(x^4+1\right)^{-1} d x$
$=\int \frac{e^{\log x^3}}{x^4+1} d x$
$=\int \frac{x^3}{x^4+1} d x \quad \ldots\left[\because e^{\log N}=N\right]$
$=\frac{1}{4} \int \frac{4 x^3}{x^4+1} d x$
$=\frac{1}{4} \int \frac{\frac{d}{d x}\left(x^4+1\right)}{x^4+1} d x$
$=\frac{1}{4} \log \left|x^4+1\right|+c . \ldots\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right]$
View full question & answer→Question 692 Marks
Integrate the following functions w.r.t. x:
$x^9 \cdot \sec ^2\left(x^{10}\right)$
AnswerLet $I=\int x^9 \cdot \sec ^2\left(x^{10}\right) d x$
Put $x^{10}=t \quad \therefore 10 x^9 d x=d t \quad \therefore x^9 d x=\frac{1}{10} d t$
$\therefore I=\int \sec ^2 t \cdot \frac{d t}{10}$
$=\frac{1}{10} \int \sec ^2 t d t$
$=\frac{1}{10} \tan t+c$
$=\frac{1}{10} \tan \left(x^{10}\right)+c .$
View full question & answer→Question 702 Marks
Integrate the following functions w.r.t. x: $\frac{1}{4 x+5 x^{-11}}$
AnswerLet $I=\int \frac{1}{4 x+5 x^{-11}} d x$$=\int \frac{x^{11}}{x^{11}\left(4 x+5 x^{-11}\right)} d x$
$=\int \frac{x^{11}}{4 x^{12}+5} d x$
$=\frac{1}{48} \int \frac{48 x^{11}}{4 x^{12}+5} d x$
$=\frac{1}{48} \int \frac{\frac{d}{d x}\left(4 x^{12}+5\right)}{4 x^{12}+5} d x$
$=\frac{1}{48} \log \left|4 x^{12}+5\right|+c \ldots\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right]$
View full question & answer→Question 712 Marks
Integrate the following functions w.r.t. x:
$\frac{e^{2 x}+1}{e^{2 x}-1}$
AnswerLet $I=\int \frac{e^{2 x}+1}{e^{2 x}-1} d x=\int \frac{\left(\frac{e^{2 x}+1}{e^x}\right)}{\left(\frac{e^{2 x}-1}{e^x}\right)} d x$
$=\int\left(\frac{e^x+e^{-x}}{e^x-e^{-x}}\right) d x$
$=\int \frac{\frac{d}{d x}\left(e^x-e^{-x}\right)}{e^x-e^{-x}} d x$
$=\log \left|e^x-e^{-x}\right|+c \quad \cdots\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right]$
View full question & answer→Question 722 Marks
Integrate the following functions w.r.t. x: $\frac{e^x \cdot \log \left(\sin e^x\right)}{\tan \left(e^x\right)}$
AnswerLet $I=\int \frac{e^x \cdot \log \left(\sin e^x\right)}{\tan \left(e^x\right)} d x$$=\int \log \left(\sin e^x\right) \cdot e^x \cot \left(e^x\right) d x$
Put $\log \left(\sin e^x\right)=t \quad \therefore \frac{1}{\sin \left(e^x\right)} \cdot \cos \left(e^x\right) \cdot e^x d x=d t$
$\therefore e^x \cdot \cot \left(e^x\right) d x=d t$
$\begin{aligned} \therefore I & =\int t d t=\frac{t^2}{2}+c \\ & =\frac{1}{2}\left[\log \left(\sin e^x\right)\right]^2+c .\end{aligned}$
View full question & answer→Question 732 Marks
Integrate the following functions w.r.t. x: $\frac{\left(x^2+2\right)}{\left(x^2+1\right)} \cdot a^{x+\tan ^{-1} x}$
AnswerLet $I=\int \frac{x^2+2}{\left(x^2+1\right)} \cdot a^{x+\tan ^{-1} x} d x$$=\int a^{x+\tan ^{-1} x} \cdot\left(\frac{x^2+2}{x^2+1}\right) d x$
Put $x+\tan ^{-1} x=t$
$\begin{array}{c}\therefore\left(1+\frac{1}{1+x^2}\right) d x=d t \\ \therefore\left(\frac{1+x^2+1}{1+x^2}\right) d x=d t \\ \therefore\left(\frac{x^2+2}{x^2+1}\right) d x=d t \\ \therefore I=\int a^t d t=\frac{a^t}{\log a}+c \\ =\frac{a^{x+\tan ^{-1} x}}{\log a}+c .\end{array}$
View full question & answer→Question 742 Marks
Integrate the following functions w.r.t. x:
$\frac{x \cdot \sec ^2\left(x^2\right)}{\sqrt{\tan ^3\left(x^2\right)}}$
AnswerLet $I=\int \frac{x \cdot \sec ^2\left(x^2\right)}{\sqrt{\tan ^3\left(x^2\right)}} d x$
Put $\tan \left(x^2\right)=t \quad \therefore \sec ^2\left(x^2\right) \times 2 x d x=d t$
$\therefore x \cdot \sec ^2\left(x^2\right) d x=\frac{d t}{2}$
$\therefore I=\int \frac{1}{\sqrt{t^3}} \cdot \frac{d t}{2}=\frac{1}{2} \int t^{-\frac{3}{2}} d t$
$=\frac{1}{2} \cdot \frac{t^{-\frac{1}{2}}}{-1 / 2}+c=\frac{-1}{\sqrt{t}}+c=\frac{-1}{\sqrt{\tan \left(x^2\right)}}+c$.
View full question & answer→Question 752 Marks
Integrate the following functions w.r.t. x:
$\frac{1+x}{x-\sin (x+\log x)}$
Answer$ \text { Let } I=\int \frac{1+x}{x \cdot \sin (x+\log x)} d x$
$= \int \frac{1}{\sin (x+\log x)} \cdot\left(\frac{1+x}{x}\right) d x$
$= \int \frac{1}{\sin (x+\log x)} \cdot\left(\frac{1}{x}+1\right) d x$
Put $x+\log x=t \quad \therefore\left(1+\frac{1}{x}\right) d x=d t$
$\therefore I=\int \frac{1}{\sin t} d t=\int \operatorname{cosec} t d t$
$=\log |\operatorname{cosec} t-\cot t|+c$
$=\log |\operatorname{cosec}(x+\log x)-\cot (x+\log x)|+c .$
View full question & answer→Question 762 Marks
Integrate the following functions w.r.t x: $\tan 3 x \tan 2 x \tan x$
View full question & answer→Question 772 Marks
Integrate the following functions w.r.t x: $\cos ^7 x$
View full question & answer→Question 782 Marks
Integrate the following functions w.r.t x: $\tan ^5 x$
View full question & answer→Question 792 Marks
Integrate the following functions w.r.t x: $\cos ^8 x \cdot \cot x$
View full question & answer→Question 802 Marks
Integrate the following functions w.r.t x: $\frac{3 e^{2 x}+5}{4 e^{2 x}-5}$
View full question & answer→Question 812 Marks
Integrate the following functions w.r.t x: $\frac{20+12 e^x}{3 e^x+4}$
View full question & answer→Question 822 Marks
Integrate the following functions w.r.t x: $\frac{\cos x}{\sin (x-a)}$
View full question & answer→Question 832 Marks
Integrate the following functions w.r.t x: $\frac{\cos 3 x-\cos 4 x}{\sin 3 x+\sin 4 x}$
View full question & answer→Question 842 Marks
Evaluate:
If $f^{\prime}(x)=x-\frac{3}{x^3}, f(1)=\frac{11}{2}$, find $f(x)$.
AnswerBy the definition of integral,
$f(x)=\int f^{\prime}(x) d x$
$=\int\left(x-\frac{3}{x^3}\right) d x$
$=\int x d x-3 \int x^{-3} d x$
$=\frac{x^2}{2}-\frac{3 x^{(-2)}}{(-2)}+c$
$=\frac{x^2}{2}+\frac{3}{2 x^2}+c$
........(1)
$f(1)=\frac{11}{2}$ ....... (Given)
$\therefore \frac{1}{2}+\frac{3}{2}+c=\frac{11}{2}$
$\therefore c=\frac{7}{2}$
$\therefore f(x)=\frac{x^2}{2}+\frac{3}{2 x^2}+\frac{7}{2} . \quad \ldots \ldots[ By (1)]$
View full question & answer→Question 852 Marks
Evaluate:
$\int \frac{3}{\sqrt{7 x-2}-\sqrt{7 x-5}} \cdot d x$
Answer$\int \frac{3}{\sqrt{7 x-2}-\sqrt{7 x-5}} d x$
$=\int \frac{3}{\sqrt{7 x-2}-\sqrt{7 x-5}} \times \frac{\sqrt{7 x-2}+\sqrt{7 x-5}}{\sqrt{7 x-2}-\sqrt{7 x-5}} d x$
$=\int \frac{3(\sqrt{7 x-2}+\sqrt{7 x-5})}{(7 x-2)-(7 x-5)} d x$
$=\int(\sqrt{7 x-2}+\sqrt{7 x-5}) d x$
$=\int(7 x-2)^{\frac{1}{2}} d x+\int(7 x-5)^{\frac{1}{2}} d x$
$=\frac{(7 x-2)^{\frac{3}{2}}}{3 / 2} \times \frac{1}{7}+\frac{(7 x-5)^{\frac{3}{2}}}{3 / 2} \times \frac{1}{7}+c$
$=\frac{2}{21}(7 x-2)^{\frac{3}{2}}+\frac{2}{21}(7 x-5)^{\frac{2}{2}}+c .$
View full question & answer→Question 862 Marks
Evaluate:
$\int \frac{2}{\sqrt{x}-\sqrt{x+3}} \cdot d x$
Answer$ \int \frac{2}{\sqrt{x}-\sqrt{x+3}} d x$
$=\int \frac{2}{\sqrt{x}-\sqrt{x+3}} \times \frac{\sqrt{x}+\sqrt{x+3}}{\sqrt{x}+\sqrt{x+3}} d x$
$=\int \frac{2(\sqrt{x}+\sqrt{x+3})}{x-(x+3)} d x$
$=-\frac{2}{3} \int(\sqrt{x}+\sqrt{x+3}) d x$
$=-\frac{2}{3} \int x^{\frac{1}{2}} d x-\frac{2}{3} \int(x+3)^{\frac{1}{2}} d x$
$=-\frac{2}{3} \cdot \frac{x^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}-\frac{2}{3} \cdot \frac{(x+3)^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}+c$
$=-\frac{4}{9}\left[x^{\frac{3}{2}}+(x+3)^{\frac{3}{2}}\right]+c .$
View full question & answer→Question 872 Marks
Evaluate:
$\int \cos ^2 x \cdot d x$
AnswerRecall the identity $\cos 2 x=2 \cos ^2 x-1$,
which gives$\cos ^2 x=\frac{1+\cos 2 x}{2}$
Therefore, $\int \cos ^2 x \cdot d x$
$=\frac{1}{2} \int(1+\cos 2 x) \cdot d x$
$=\frac{1}{2} \int d x+\frac{1}{2} \int \cos 2 x \cdot d x$
$=\frac{x}{2}+\frac{1}{4} \sin 2 x+C .$
View full question & answer→Question 882 Marks
Evaluate:$\int \sqrt{1+\sin 5 x} \cdot d x$
Answer$\begin{aligned} & \int \sqrt{1+\sin 5 x} \cdot d x \\ = & \int \sqrt{\sin ^2 x+\cos ^2 x+5 \sin x \cos x} \cdot d x \\ = & \int \sqrt{(\sin x+\cos x)^2} \cdot d x \\ = & \int(\sin x+\cos x) \cdot d x \\ = & \int \sin x d x+\int \cos x \cdot d x \\ = & \left(\frac{2}{5} \sin \frac{5 x}{2}-\cos \frac{5 x}{2}\right)+c .\end{aligned}$
View full question & answer→Question 892 Marks
Evaluate:
$\int \frac{\sin 4 x}{\cos 2 x} \cdot d x$
Answer$\int \frac{\sin 4 x}{\cos 2 x} d x$
$=\int \frac{2 \sin 2 x \cos 2 x}{\cos 2 x} d x$
$=2 \int \sin 2 x d x$
$=2\left(-\frac{\cos 2 x}{2}\right)+c$
$=-\cos 2 x+c .$
View full question & answer→Question 902 Marks
Evaluate:
$\int \frac{2 x-7}{\sqrt{4 x-1}} \cdot d x$
Answer$ \int \frac{2 x-7}{\sqrt{4 x-1}} d x$
$= \frac{1}{2} \int \frac{2(2 x-7)}{\sqrt{4 x-1}} d x$
$= \frac{1}{2} \int \frac{(4 x-1)-13}{\sqrt{4 x-1}} d x$
$=\frac{1}{2} \int\left(\frac{4 x-1}{\sqrt{4 x-1}}-\frac{13}{\sqrt{4 x-1}}\right) d x$
$=\frac{1}{2} \int(4 x-1)^{\frac{1}{2}} d x-\frac{13}{2} \int(4 x-1)^{-\frac{1}{2}} d x$
$=\frac{1}{2} \cdot \frac{(4 x-1)^{\frac{3}{2}}}{(4)\left(\frac{3}{2}\right)}-\frac{13}{2} \cdot \frac{(4 x-1)^{\frac{1}{2}}}{(4)\left(\frac{1}{2}\right)}+c$
$=\frac{1}{12}(4 x-1)^{\frac{3}{2}}-\frac{13}{4} \sqrt{4 x-1}+c$.
View full question & answer→Question 912 Marks
Evaluate:$\int \frac{x-2}{\sqrt{x+5}} \cdot d x$
Answer$\begin{aligned} & \int \frac{x-2}{\sqrt{x+5}} d x=\int \frac{(x+5)-7}{\sqrt{x+5}} d x \\ = & \int\left(\frac{x+5}{\sqrt{x+5}}-\frac{7}{\sqrt{x+5}}\right) d x \\ = & \int(x+5)^{\frac{1}{2}} d x-7 \int(x+5)^{-\frac{1}{2}} d x \\ = & \frac{(x+5)^{\frac{3}{2}}}{(3 / 2)}-\frac{7(x+5)^{\frac{1}{2}}}{(1 / 2)}+c\end{aligned}$$=\frac{2}{3}(x+5)^{\frac{3}{2}}-14 \sqrt{x+5}+c$.
View full question & answer→Question 922 Marks
Evaluate:
$\int \frac{5 x+2}{3 x-4} \cdot d x$
Answer$\int \frac{5 x+2}{3 x-4} d x$
$=\int \frac{\frac{5}{3}(3 x-4)+\frac{20}{3}+2}{3 x-4} d x$
$=\int \frac{\frac{5}{3}(3 x-4)+\frac{26}{3}}{3 x-4} d x$
$=\int\left[\frac{5}{3}+\frac{\left(\frac{26}{3}\right)}{3 x-4}\right] d x$
$=\frac{5}{3} \int 1 d x+\frac{26}{3} \int \frac{1}{3 x-4} d x$
$=\frac{5 x}{3}+\frac{26}{3} \cdot \frac{1}{3} \log |3 x-4|+c$
$=\frac{5 x}{3}+\frac{26}{9} \log |3 x-4|+c .$
View full question & answer→Question 932 Marks
Evaluate:
$\int \frac{4 x+3}{2 x+1} \cdot d x$
Answer$ \int \frac{4 x+3}{2 x+1} d x $
$ = \int \frac{2(2 x+1)+1}{2 x+1} d x$
$=\int\left(\frac{2(2 x+1)}{2 x+1}+\frac{1}{2 x+1}\right) d x$
$=2 \int 1 d x+\int \frac{1}{2 x+1} d x$
$=2 x+\frac{1}{2} \log |2 x+1|+c .$
View full question & answer→Question 942 Marks
Evaluate:
$\int \frac{x}{x+2} \cdot d x$
Answer$\quad \int \frac{x}{x+2} d x$
$=\int \frac{(x+2)-2}{x+2} d x$
$=\int\left(\frac{x+2}{x+2}-\frac{2}{x+2}\right) d x$
$=\int 1 d x-2 \int \frac{1}{x+2} d x$
$=x-2 \log |x+2|+c .$
View full question & answer→