Mathematics - Definite Integration Question with Solution | TestHub

Let, $I={\int }_0^{100\pi }\frac{{\sin }^{2}x}{e^{\left({\displaystyle \frac{x}{\pi }-\left[\frac{x}{\pi }\right]}\right)}}dx$

We know that if $f\left(x\right)$ is a periodic function, with period $T$ then ${\int }_0^{nT}f\left(x\right)dx=n{\int }_0^{T}f\left(x\right)dx, n\in Z$

$\Rightarrow I=100{\int }_0^{\pi }\frac{{\sin }^{2}x}{e^{\left({\displaystyle \frac{x}{\pi }-\left[\frac{x}{\pi }\right]}\right)}}dx$

For, $0<x<\pi ,0<\frac{x}{\pi }<1$ $\Rightarrow \left[\frac{x}{\pi }\right]=0$

$\Rightarrow I=100{\int }_0^{\pi }\frac{{\sin }^{2}x}{e^{\left({\displaystyle \frac{x}{\pi }}\right)}}dx$

Now, using $\cos 2x=1-2{\sin }^{2}x,$ we get

$I=100{\int }_0^{\pi }{e}^{-x/\pi }\frac{\left(1-\cos 2x\right)}{2}dx$

$\Rightarrow I=100\times \frac{1}{2}{\int }_0^{\pi }\left(e^{-x/\pi }-e^{-x/\pi }·\cos 2x\right)dx$

$\Rightarrow I=50\left\{{\int }_0^{\pi }{e}^{-x/\pi }dx-{\int }_0^{\pi }{e}^{-x/\pi }·\cos 2xdx\right\}$

Let, $I_1={\int }_0^{\pi }{e}^{-x/\pi }dx$

$\Rightarrow I_1={\left[-\pi e^{-x/\pi }\right]}_0^{\pi }$

$\Rightarrow I_1=-\pi \left(e^{-\mathrm{\pi }/\pi }-e^0\right)=\pi \left(1-e^{-1}\right)$

And, let $I_2={\int }_0^{\pi }{e}^{-x/\pi }·\cos 2xdx$

By using $ILATE$ rule,

$I_2={\int }_0^{\pi }\cos 2x·e^{-x/\pi }dx$

Now, using integration by parts, i.e. $\int \left(u·v\right)dx=u\int vdx-\int \left[\frac{d}{dx}\left(u\right)\int vdx\right]dx,$ we get

$\Rightarrow I_2=-\pi {\left[{\mathrm{e}}^{-x/\pi }·\cos 2x\right]}_0^{\pi }-{\int }_0^{\mathrm{\pi }}\left(-\pi {\mathrm{e}}^{-x/\pi }·\left(-2\sin 2x\right)\right)dx$

$\Rightarrow I_2=\pi \left(1-e^{-1}\right)-2\pi {\int }_0^{\pi }{e}^{-x/\pi }\sin 2xdx$

Again, applying integration by parts, we get

$\left.I_2=\pi \left(1-e^{-1}\right)-2\pi \{{\left[-\pi e^{-x/\pi }\sin 2x\right]}_0^{\pi }-{\int }_0^{\pi }-\pi e^{-x/\pi }2\cos 2xdx\right\}$

$\Rightarrow I_2=\pi \left(1-e^{-1}\right)-4{\pi }^2{I}_2$

$\Rightarrow I_2=\frac{\pi \left(1-e^{-1}\right)}{1+4{\pi }^2}$

$\therefore I=50\left(I_1+I_2\right)$

$\Rightarrow I=50\left\{\pi \left(1-e^{-1}\right)-\frac{\pi \left(1-e^{-1}\right)}{1+4{\pi }^2}\right\}$

$\Rightarrow I=\frac{200\left(1-e^{-1}\right){\pi }^3}{1+4{\pi }^2}$

Given $I=\frac{\alpha {\pi }^3}{1+4{\pi }^2}$

$\Rightarrow \frac{\alpha {\pi }^3}{1+4{\pi }^2}=\frac{200\left(1-e^{-1}\right){\pi }^3}{1+4{\pi }^2}$

$\Rightarrow \alpha =200\left(1-e^{-1}\right).$