Solution

$\int_0^{1} \dfrac{x}{x^{2}+1} d x$

Let $x^{2}+1=t \Rightarrow 2 x d x=d t$

When $x=0, t=1$ and when $x=1, t=2$

$\therefore \int_0^{1} \dfrac{x}{x^{2}+1} d x=\dfrac{1}{2} \int_1^{2} \dfrac{d t}{t}$

$=\dfrac{1}{2}[\log |t|]_1^{2}$

$=\dfrac{1}{2}[\log 2-\log 1]$

$=\dfrac{1}{2} \log 2$

Solution

Let $I=\int_0^{\dfrac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi d \phi=\int_0^{\dfrac{\pi}{2}} \sqrt{\sin \phi} \cos ^{4} \phi \cos \phi d \phi$

Also, let $\sin \phi=t \Rightarrow \cos \phi d \phi=d t$

When $\phi=0, t=0$ and when $\phi=\dfrac{\pi}{2}, t=1$

$\therefore I=\displaystyle\int_0^{1} \sqrt{t}(1-t^{2})^{2} d t$

$=\displaystyle\int_0^{1} t^{\dfrac{1}{2}}(1+t^{4}-2 t^{2}) d t$

$=\displaystyle\int_0^{1}[t^{\dfrac{1}{2}}+t^{\dfrac{9}{2}}-2 t^{\dfrac{5}{2}}] d t$

$=\Bigg[\dfrac{t^{\dfrac{3}{2}}}{\dfrac{3}{2}}+\dfrac{t^{\dfrac{11}{2}}}{\dfrac{11}{2}}-\dfrac{2 t^{\dfrac{7}{2}}}{\dfrac{7}{2}}\Bigg]_0^{1}$

$=\dfrac{2}{3}+\dfrac{2}{11}-\dfrac{4}{7}$

$=\dfrac{154+42-132}{231}$

$=\dfrac{64}{231}$

Solution

Let $I=\int_0^{1} \sin ^{-1}(\dfrac{2 x}{1+x^{2}}) d x$

Also, let $x=\tan \theta \Rightarrow d x=\sec ^{2} \theta d \theta$

When $x=0, \theta=0$ and when $x=1, \theta=\dfrac{\pi}{4}$

$ \begin{aligned} I & =\int_0^{\pi/4} \sin ^{-1}(\dfrac{2 \tan \theta}{1+\tan ^{2} \theta}) \sec ^{2} \theta d \theta \\ & =\int_0^{\dfrac{\pi}{4}} \sin ^{-1}(\sin 2 \theta) \sec ^{2} \theta d \theta \\ & =\int_0^{\dfrac{\pi}{4}} 2 \theta \cdot \sec ^{2} \theta d \theta \\ & =2 \int_0^{\pi/4} \theta \cdot \sec ^{2} \theta d \theta \end{aligned} $

Taking $\theta$ as first function and $\sec ^{2} \theta$ as second function and integrating by parts, we obtain

$ \begin{aligned} I & =2\Bigg[\theta \int \sec ^{2} \theta d \theta-\int{(\dfrac{d}{d x} \theta) \int \sec ^{2} \theta d \theta} d \theta\Bigg]_0^{\dfrac{\pi}{4}} \\ & =2\Bigg[\theta \tan \theta-\int \tan \theta d \theta\Bigg]_0^{\dfrac{\pi}{4}} \\ & =2\Bigg[\theta \tan \theta+\log |\cos \theta|\Bigg]_0^{\dfrac{\pi}{4}} \\ & =2[\dfrac{\pi}{4} \tan \dfrac{\pi}{4}+\log |\cos \dfrac{\pi}{4}|-\log |\cos 0|] \\ & =2[\dfrac{\pi}{4}+\log (\dfrac{1}{\sqrt{2}})-\log 1] \\ & =2[\dfrac{\pi}{4}-\dfrac{1}{2} \log 2] \\ & =\dfrac{\pi}{2}-\log 2 \end{aligned} $

Solution

$\int_0^{2} x \sqrt{x+2} d x$

Let $x+2=t^{2} \Rightarrow d x=2 t d t$

When $x=0, t=\sqrt{2}$ and when $x=2, t=2$

$\therefore \displaystyle\int_0^{2} x \sqrt{x+2} d x=\int _{\sqrt{2}}^{2}(t^{2}-2) \sqrt{t^{2}} 2 t\ d t$

$=2 \displaystyle\int _{\sqrt{2}}^{2}(t^{2}-2)t^{2} d t$

$=2 \displaystyle\int _{\sqrt{2}}^{2}(t^{4}-2 t^{2}) d t$

$=2\Bigg[\dfrac{t^{5}}{5}-\dfrac{2 t^{3}}{3}\Bigg] _{\sqrt{2}}^{2}$

$=2\Bigg[\dfrac{32}{5}-\dfrac{16}{3}-\dfrac{4 \sqrt{2}}{5}+\dfrac{4 \sqrt{2}}{3}\Bigg]$

$=2\Bigg[\dfrac{96-80-12 \sqrt{2}+20 \sqrt{2}}{15}\Bigg]$

$=2\Bigg[\dfrac{16+8 \sqrt{2}}{15}\Bigg]$

$=\dfrac{16(2+\sqrt{2})}{15}$

$=\dfrac{16 \sqrt{2}(\sqrt{2}+1)}{15}$

Solution

$\displaystyle\int_0^{\dfrac{\pi}{2}} \dfrac{\sin x}{1+\cos ^{2} x} d x$

Let $\cos x=t \Rightarrow -\sin x d x=d t$

When $x=0, t=1$ and when $x=\dfrac{\pi}{2}, t=0$

$\Rightarrow \displaystyle\int_0^{\dfrac{\pi}{2}} \dfrac{\sin x}{1+\cos ^{2} x} d x=-\int_0^{1} \dfrac{d t}{1+t^{2}}$

$=-[\tan ^{-1} t]_1^{0}$

$=-[\tan ^{-1} 0-\tan ^{-1} 1]$

$=-[-\dfrac{\pi}{4}]$

$=\dfrac{\pi}{4}$

Solution

$\displaystyle\int_0^{2} \dfrac{d x}{x+4-x^{2}}=\int_0^{2} \dfrac{d x}{-(x^{2}-x-4)}$

$=\displaystyle\int_0^{2} \dfrac{d x}{-(x^{2}-x+\dfrac{1}{4}-\dfrac{1}{4}-4)}$

$=\displaystyle\int_0^{2} \dfrac{d x}{-[(x-\dfrac{1}{2})^{2}-\dfrac{17}{4}]}$

$=\displaystyle\int_0^{2} \dfrac{d x}{(\dfrac{\sqrt{17}}{2})^{2}-(x-\dfrac{1}{2})^{2}}$

Let $\quad x-\dfrac{1}{2}=t \Rightarrow d x=d t$

When $x=0, t=-\dfrac{1}{2}$ and when $x=2, t=\dfrac{3}{2}$

$\therefore \displaystyle\int_0^{2} \dfrac{d x}{(\dfrac{\sqrt{17}}{2})^{2}-(x-\dfrac{1}{2})^{2}}=\displaystyle \int_{-\dfrac{1}{2}}^{\dfrac{3}{2}} \dfrac{d t}{(\dfrac{\sqrt{17}}{2})^{2}-t^{2}}$

$=\Bigg[\dfrac{1}{2(\dfrac{\sqrt{17}}{2})} \log \dfrac{\dfrac{\sqrt{17}}{2}+t}{\dfrac{\sqrt{17}}{2}-t}\Bigg] _{-\dfrac{1}{2}}^{\dfrac{3}{2}}$

$=\dfrac{1}{\sqrt{17}}\Bigg[\log \dfrac{\dfrac{\sqrt{17}}{2}+\dfrac{3}{2}}{\dfrac{\sqrt{17}}{2}-\dfrac{3}{2}}-\dfrac{\log \dfrac{\sqrt{17}}{2}-\dfrac{1}{2}}{\log \dfrac{\sqrt{17}}{2}+\dfrac{1}{2}}\Bigg]$

$=\dfrac{1}{\sqrt{17}}\Bigg[\log \dfrac{\sqrt{17}+3}{\sqrt{17}-3}-\log \dfrac{\sqrt{17}-1}{\sqrt{17}+1}\Bigg]$

$=\dfrac{1}{\sqrt{17}} \log \dfrac{\sqrt{17}+3}{\sqrt{17}-3} \times \dfrac{\sqrt{17}+1}{\sqrt{17}-1}$

$=\dfrac{1}{\sqrt{17}} \log [\dfrac{17+3+4 \sqrt{17}}{17+3-4 \sqrt{17}}]$

$=\dfrac{1}{\sqrt{17}} \log [\dfrac{20+4 \sqrt{17}}{20-4 \sqrt{17}}]$

$=\dfrac{1}{\sqrt{17}} \log (\dfrac{5+\sqrt{17}}{5-\sqrt{17}})$

$=\dfrac{1}{\sqrt{17}} \log [\dfrac{(5+\sqrt{17})(5+\sqrt{17})}{25-17}]$

$=\dfrac{1}{\sqrt{17}} \log [\dfrac{25+17+10 \sqrt{17}}{8}]$

$=\dfrac{1}{\sqrt{17}} \log (\dfrac{42+10 \sqrt{17}}{8})$

$=\dfrac{1}{\sqrt{17}} \log (\dfrac{21+5 \sqrt{17}}{4})$

Solution

$\displaystyle\int _{-1}^{1} \dfrac{d x}{x^{2}+2 x+5}=\int _{-1}^{1} \dfrac{d x}{(x^{2}+2 x+1)+4}=\int _{-1}^{1} \dfrac{d x}{(x+1)^{2}+(2)^{2}}$

Let $x+1=t \Rightarrow d x=d t$

When $x=-1, t=0$ and when $x=1, t=2$

$\therefore \displaystyle\int _{-1}^{1} \dfrac{d x}{(x+1)^{2}+(2)^{2}}=\int_0^{2} \dfrac{d t}{t^{2}+2^{2}}$

$=\Bigg[\dfrac{1}{2} \tan ^{-1} \dfrac{t}{2}\Bigg]_0^{2}$

$=\dfrac{1}{2} \tan ^{-1} 1-\dfrac{1}{2} \tan ^{-1} 0$

$=\dfrac{1}{2}(\dfrac{\pi}{4})=\dfrac{\pi}{8}$

Solution

$\int(\dfrac{1}{x}-\dfrac{1}{2 x^{2}}) e^{2 x} d x$

Let $2 x=t \Rightarrow 2 d x=d t$

When $x=1, t=2$ and when $x=2, t=4$

$ \begin{aligned} \therefore \int_1^{2}(\dfrac{1}{x}-\dfrac{1}{2 x^{2}}) e^{2 x} d x & =\dfrac{1}{2} \int_2^{4}(\dfrac{2}{t}-\dfrac{2}{t^{2}}) e^{t} d t \\ & =\int_2^{4}(\dfrac{1}{t}-\dfrac{1}{t^{2}}) e^{t} d t \end{aligned} $

Let $\dfrac{1}{t}=f(t)$

Then, $f^{\prime}(t)=-\dfrac{1}{t^{2}}$

$ \begin{aligned} \Rightarrow \int_2^{4}(\dfrac{1}{t}-\dfrac{1}{t^{2}}) e^{t} d t & =\int_2^{4} e^{t}[f(t)+f^{\prime}(t)] d t \\ & =[e^{t} f(t)]_2^{4} \\ & =[e^{t} \cdot \dfrac{2}{t}]_2^{4} \\ & =[\dfrac{e^{t}}{t}]_2^{4} \\ & =\dfrac{e^{4}}{4}-\dfrac{e^{2}}{2} \\ & =\dfrac{e^{2}(e^{2}-2)}{4} \end{aligned} $

Solution

Let $I=\displaystyle\int _{\dfrac{1}{3}}^{1} \dfrac{(x-x^{3})^{\dfrac{1}{3}}}{x^{4}} d x$

Also, let $x=\sin \theta \Rightarrow d x=\cos \theta d \theta$

When $x=\dfrac{1}{3}, \theta=\sin ^{-1}(\dfrac{1}{3})$ and when $x=1, \theta=\dfrac{\pi}{2}$

$\Rightarrow I=\int _{\sin ^{-1}(\dfrac{1}{3})}^{\dfrac{\pi}{2}} \dfrac{(\sin \theta-\sin ^{3} \theta)^{\dfrac{1}{3}}}{\sin ^{4} \theta} \cos \theta d \theta$

$ =\int _{\sin ^{-1}(\dfrac{1}{3})}^{\dfrac{\pi}{2}} \dfrac{(\sin \theta)^{\dfrac{1}{3}}(1-\sin ^{2} \theta)^{\dfrac{1}{3}}}{\sin ^{4} \theta} \cos \theta d \theta $

$=\int _{\sin ^{-1}(\dfrac{1}{3})}^{\dfrac{\pi}{2}} \dfrac{(\sin \theta)^{\dfrac{1}{3}}(\cos \theta)^{\dfrac{2}{3}}}{\sin ^{4} \theta} \cos \theta d \theta$

$=\int _{\sin ^{-1}(\dfrac{1}{3})}^{\dfrac{\pi}{2}} \dfrac{(\sin \theta)^{\dfrac{1}{3}}(\cos \theta)^{\dfrac{2}{3}}}{\sin ^{2} \theta \sin ^{2} \theta} \cos \theta d \theta$

$=\int _{\sin ^{-1}(\dfrac{1}{3})}^{\dfrac{\pi}{2}} \dfrac{(\cos \theta)^{\dfrac{5}{3}}}{(\sin \theta)^{\dfrac{5}{3}}} cosec^{2} \theta d \theta$

$=\int _{\sin ^{-1}(\dfrac{1}{3})}^{\dfrac{\pi}{2}}(\cot \theta)^{\dfrac{5}{3}} cosec^{2} \theta d \theta$

Let $\cot \theta=t \Rightarrow -cosec 2 \theta d \theta=d t$

$ \begin{aligned} \therefore I & =-\int _{2 \sqrt{2}}^{0}(t)^{\dfrac{5}{3}} d t \\ & =-\Bigg[\dfrac{3}{8}(t)^{\dfrac{8}{3}}\Bigg] _{2 \sqrt{2}}^{0} \\ & =-\dfrac{3}{8}\Bigg[(t)^{\dfrac{8}{3}}\Bigg] _{2 \sqrt{2}}^{0} \\ & =-\dfrac{3}{8}[-(2 \sqrt{2})^{\dfrac{8}{3}}] \\ & =\dfrac{3}{8}[(\sqrt{8})^{\dfrac{8}{3}}] \\ & =\dfrac{3}{8}[(8)^{\dfrac{4}{3}}] \\ & =\dfrac{3}{8}[16] \\ & =3 \times 2 \\ & =6 \end{aligned} $

$ \text{ When } \theta=\sin ^{-1}(\dfrac{1}{3}), t=2 \sqrt{2} \text{ and when } \theta=\dfrac{\pi}{2}, t=0 $

Hence, the correct Answer is A.

Solution

$f(x)=\int_0^{x} t \sin t d t$

Integrating by parts, we obtain

$ \begin{aligned} f(x) & =t \int_0^{x} \sin t d t-\int_0^{x}{\dfrac{d}{d t} (t) \int \sin t d t} d t \\ & =[t(-\cos t)]_0^{x}-\int_0^{x}(-\cos t) d t \\ & =[-t \cos t+\sin t]_0^{x} \\ & =-x \cos x+\sin x \\ \Rightarrow f^{\prime}(x) & =-[{x(-\sin x)}+\cos x]+\cos x \\ & =x \sin x-\cos x+\cos x \\ & =x \sin x \end{aligned} $

Hence, the correct Answer is B.