Integrals

$\begin{array}{l}Given,f\left(x\right)={\displaystyle {\int }_0^{x}t}sintdt\\ =t{\displaystyle {\int }_0^{x}sintdt}-{\displaystyle {\int }_0^x\frac{dt}{dt}}{\displaystyle \int sintdtdt}\\ ={\left[t\left(-cost\right)\right]}_0^x-{\displaystyle {\int }_0^x\left(-cost\right)}dt\\ =-\left[xcosx-0cos0\right]+{\left[sint\right]}_0^x\\ =-xcosx+\left[sinx-sin0\right]\\ =-xcosx+sinx\\ \therefore f\text{'}\left(x\right)=\frac{d}{dx}\left(-xcosx+sinx\right)\\ =-x\frac{d}{dx}cosx-cosx\frac{dx}{dx}+\frac{d}{dx}sinx\\ =-x\left(-sinx\right)-cosx+sinx\\ =xsinx\end{array}$

? Option (B) is correct.

$\begin{array}{l}Let,I={\displaystyle {\int }_{1/3}^1\frac{{\left(x-x^3\right)}^{1/3}}{x^4}}dx\\ ={\displaystyle {\int }_{1/3}^1{\left[\frac{x^3\left(\frac{1}{x^2}-1\right)}{x^4}\right]}^{1/3}}dx\\ ={\displaystyle {\int }_{1/3}^1\frac{x{\left(\frac{1}{x^2}-1\right)}^{1/3}}{x^4}}dx={\displaystyle {\int }_{1/3}^1\frac{{\left(\frac{1}{x^2}-1\right)}^{1/3}}{x^3}}dx\\ ={\displaystyle {\int }_{1/3}^1{\left(x^{-2}-1\right)}^{1/3}}{x}^{-3}dx\end{array}$

= 6

Option A is correct

$\begin{array}{l}Let={\displaystyle {\int }_1^2\left(\frac{1}{x}-\frac{1}{2x^2}\right)}{e}^{2x}dx\\ Let,2x=t\Rightarrow dx=\frac{dt}{2}\\ When,x=1,t=2*1=2\\ x=2,t=2*2=4\end{array}$

$\begin{array}{l}So,I={\displaystyle {\int }_2^4\left(\frac{1}{\left(t/2\right)}-\frac{1}{t\left(\frac{t}{2}\right)}\right)}{e}^t\frac{dt}{2}\\ ={\displaystyle {\int }_2^4\left(\frac{2}{t}-\frac{2}{t^2}\right)}{e}^t\frac{dt}{2}\end{array}$

$={\displaystyle {\int }_2^4\left(\frac{1}{t}+\frac{\left(-1\right)}{t^2}\right)}{e}^{t}dt$ is in the form

$\begin{array}{l}Let,={\displaystyle {\int }_{-1}^1\frac{dx}{x^2+2x+5.}}\\ ={\displaystyle {\int }_{-1}^1\frac{dx}{x^2+2x+1+4}}\\ ={\displaystyle {\int }_{-1}^1\frac{dx}{{\left(x+1\right)}^2+4}}={\displaystyle {\int }_{-1}^1\frac{dx}{{\left(x+1\right)}^2+2^2.}}\end{array}$

Let x + 1 = t $\Rightarrow$ dx = dt

When x = 1, t = 1 + 1 = 2

x = –1, t = –1 + 1 = 0

$\begin{array}{l}I\; ={\displaystyle {\int }_0^2\frac{dt}{t^2+2^2}}\\ =\frac{1}{2}{\left[ta{n}^{-1}\frac{t}{2}\right]}_0^2\\ =\frac{1}{2}\left[ta{n}^{-1}\frac{2}{2}-ta{n}^{-1}\frac{0}{2}\right]\\ =\frac{1}{2}\left(ta{n}^{-1}1-ta{n}^{-1}0\right)\end{array}$

$=\frac{1}{2}[\pi /4$
$\begin{array}{l}\frac{\mathrm{}}{}-0]\\ =\pi /8\end{array}$

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

$\begin{array}{l}Let={\displaystyle {\int }_0^{1}si{n}^{-1}\left(\frac{2x}{1+x^2}\right)}dx\\ Putting,x=ta\mathrm{n\theta }\Rightarrow \theta =ta{n}^{-1}x\&dx=se{c}^2\mathrm{\theta d\theta }\\ When,x=0,\theta =ta{n}^{-1}\left(0\right)=0\end{array}$

Kindly go through the solution