Question

Evaluate: ${\int }_0^{\frac{\pi }{2}}\frac{xsinxcosx}{si{n}^{4}x+co{s}^{4}x}dx$

OR

Evaluate ${\int }_0^4(x+e^{2x})dx$ as the limit of a sum.

Answer 1

$I={\int }_0^{\frac{\pi }{2}}\frac{xsinxcosx}{si{n}^{4}x+co{s}^{4}x}$

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{(\frac{\pi }{2}-x)sin(\frac{\pi }{2}-x)cos(\frac{\pi }{2}-x)}{si{n}^4(\frac{\pi }{2}-x)+co{s}^4(\frac{\pi }{2}-x)}$

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{(\frac{\pi }{2}-x)cosxsinx}{co{s}^{4}x+si{n}^{4}x}dx\dots \left(ii\right)$

Adding (i) and (ii) $2I=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\frac{cos!xsinx}{co{s}^{4}x+si{n}^{4}x}dx$

$\Rightarrow 2I=\frac{\pi }{2}\times 2{\int }_0^{\frac{\pi }{2}}\frac{cosxsinx}{co{s}^{4}x+si{n}^{4}x}dx$

$\Rightarrow 2I=\frac{\pi }{2}[{\int }_0^{\frac{\pi }{2}}\frac{cosxsinx}{co{s}^{4}x+si{n}^{4}x}dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\frac{cosxsinx}{co{s}^{4}x+si{n}^{4}x}]dx$

$\Rightarrow 2I=\frac{\pi }{2}[{\int }_0^{\frac{\pi }{4}}\frac{tanxse{c}^{2}x}{1+ta{n}^{4}x}dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\frac{cotxcose{c}^{2}x}{co{t}^{4}x+1}dx]$

Substituting $ta{n}^{2}x=t$ and $co{t}^{2}x=u$ in both integrals respectively

$\Rightarrow 2tanxse{c}^{2}xdx=dt$ and $2cose{c}^{2}xcotxdx=-du$

Also when $x=0\Rightarrow t=0,$ when $x=\frac{\pi }{4}\Rightarrow t=1$

And, when $x=\frac{\pi }{4}\Rightarrow u=1$, when $x=\frac{\pi }{2}\Rightarrow u=0$

$\therefore 2I=\frac{\pi }{2}[\frac{1}{2}{\int }_0^1\frac{dt}{1+t^2}-\frac{1}{2}{\int }_1^0\frac{du}{1+u^2}]$

$\Rightarrow 2I=\frac{\pi }{2}[\frac{1}{2}{\int }_0^1\frac{dt}{1+t^2}+\frac{1}{2}{\int }_1^0\frac{du}{1+u^2}]$

$\Rightarrow I=\frac{\pi }{8}\left[\right[ta{n}^{-1}t{]}_0^1+\left[ta{n}^{-1}u{]}_0^1\right]$

$\Rightarrow I=\frac{\pi }{8}[\frac{\pi }{4}+\frac{\pi }{4}]=\frac{{\pi }^2}{16}or,I={\left(\frac{\pi }{4}\right)}^2$

OR

Let ${\int }_0^4(x+e^{2x})dx$

$\because {\int }_a^{b}f\left(x\right)dx={lim}_{n\to \infty }\left[f\right(a)+f(a+h)+f(a+2h)+\dots +f(a+(n-1)h\left)\right]$

$\therefore {\int }_a^{b}f\left(x\right)dx={lim}_{n\to \infty }{\sum }_{r=0}^{n-1}f(a+rh)[\because n\to \infty ,h\to 0\Rightarrow nh=b-a]$

Here $f\left(x\right)=x+e^{2x}$ and $a=0b=4$ so $nh=4-0=4$

$\Rightarrow f(0+rh)=0rh+e^{2\times 0+2rh}=rh+e^{2rh}$

$\therefore I={\int }_0^4(x+e^{2x})dx={lim}_{n\to \infty }{\sum }_{r=0}^{n-1}f(a+rh)[rh+e^{2h}]$

$\Rightarrow I={lim}_{n\to \infty }\{h{\sum }_{r=0}^{n-1}r+{\sum }_{r=0}^{n-1}{e}^{2rh}\}$

$\Rightarrow I={lim}_{n\to \infty }\{h\left(\frac{n(n-1)}{2}\right)+[1+e^{2h}+e^{4h}+\dots +e^{2(n-1)h}\left]\right\}$

$\Rightarrow I={lim}_{n\to \infty }\{\frac{nh(nh-h)}{2}+h\left[\frac{1\times \left[\right(e^{2h}{)}^n-1]}{e^{2h}-1}\right]\}$

$\Rightarrow I={lim}_{n\to \infty }\{\frac{nh(nhy-h)}{2}+\frac{2h}{e^{2h}-1}\times \frac{e^{2nh}-1}{2}\}$

$\Rightarrow I=\{\frac{4(4-0)}{2}+1\times \frac{e^{2\times 4}-1}{2}\}$

$\therefore I=\frac{15+e^8}{2}$