Maths Integrals

$\begin{array}{l}LetI={\displaystyle {\int }_{-5}^5|}x+2|dx\\ ={\displaystyle {\int }_{-5}^{-2}|}x+2|dx+{\displaystyle {\int }_{-2}^5|}x+2|dx\end{array}$

? |x + 2| = x + 2 if if x + 2 > 0 => x> –2

– (x + 2) if x + 2 < 0 => x< 2.

$\begin{array}{l}={\left[-\frac{x^2}{2}-2x\right]}_{-5}^{-2}+{\left[\frac{x^2}{2}+2x\right]}_{-2}^5\\ =-\left[\left(\frac{{(-2)}^2}{2}+2(-2)\right)-\left(\frac{{\left(-5\right)}^2}{2}+2(-5)\right)\right]+\left[\left(\frac{5^2}{2}+2*5\right)-\left(\frac{{(-2)}^2}{2}+2(-2)\right)\right]\\ =-\left[2-4-\frac{25}{2}+10\right]+\left[\frac{25}{2}+10-2+4\right]\\ =-8+\frac{25}{2}+\frac{25}{2}+12\\ ==–8+25+12=29.\end{array}$

$\begin{array}{l}Let,I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^{5}xdx}{si{n}^{5}x+co{s}^{5}x}}-----(i)\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^5\left(\frac{\pi }{2}-x\right)}{si{n}^5\left(\frac{\pi }{2}-x\right)+co{s}^5\left(\frac{\pi }{2}-x\right)}}dx\\ I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{si{n}^{5}x}{co{s}^{5}x+si{n}^{5}x}}dx-----(ii)\end{array}$

$\begin{array}{l}Adding\left(i\right)\&\left(ii\right)weget,\\ 2I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left\{\frac{co{s}^{5}x}{si{n}^{5}x+co{s}^{5}x}+\frac{si{n}^{5}x}{co{s}^{5}x+si{n}^{5}x}\right\}}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^{5}x+si{n}^{5}x}{si{n}^{5}x+co{s}^{3}x}}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}d}x=\frac{\pi }{2}\\ I=\frac{\pi }{4}.\end{array}$

$\begin{array}{l}Let,I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x+co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}}dx-----(i)\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{\pi }{2}-x\right)}{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{\pi }{2}-x\right)+co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{\pi }{2}-x\right)}}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}{co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x+si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}}dx-----(ii)\\ \left(i\right)+\left(ii\right)\\ 2I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left\{\frac{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x+co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}+\frac{co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}{co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x+si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}\right\}}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left\{\frac{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x+co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x+co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}\right\}}dx\\ 2I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}d}x=\frac{\pi }{2}\\ I=\frac{\mathrm{\pi /4}}{}\end{array}$

Kindly go through the solution

$LetI={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}co{s}^2}xdx·-----(i)$

$\begin{array}{l}={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}cos{}^2}\left(\frac{\pi }{2}-x\right)dx\{?{\displaystyle {\int }_0^{a}f}(x)={\displaystyle {\int }_0^{a}f}(a-x)dx.\\ I={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}si{n}^{2}xdx}-----(ii)\\ Adding\left(i\right)\&\left(ii\right),\\ 2I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(co{s}^{2}x+si{n}^{2}x\right)}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}1}·dx\\ 2I={[x]}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{\pi }{2}-0\\ 2I=\frac{\pi }{2}\\ I=\frac{\pi }{}\end{array}$

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

? 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