Question

Evaluating ${\int }_0^{\frac{\pi }{2}}\frac{co{s}^{2}xdx}{1+3si{n}^{2}x}$

Answer 1

Let I = ${\int }_0^{\frac{\pi }{2}}\frac{co{s}^{2}xdx}{1+3si{n}^{2}x}\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{se{c}^{2}xdx}{se{c}^{4}x+3se{c}^{2}xta{n}^{2}x}\text{on Dividing Nr \& Dr by}co{s}^{4}x$

$\therefore I={\int }_0^{\frac{pi}{2}}\frac{se{c}^{2}xdx}{se{c}^{2}x(1+ta{n}^{2}x+3ta{n}^{2}x)}\Rightarrow I={\int }_0^{\frac{pi}{2}}\frac{se{c}^{2}xdx}{(1+ta{n}^{2}x)(1+4ta{n}^{2}x)}$

Put tan x = t $\Rightarrow se{c}^2$xdx = dt. When x = 0 $Rightarrow$ t = 0 & when x = $\frac{\pi }{2}\Rightarrow t=\infty$

So I = ${\int }_0^{\infty }\frac{dt}{(1+t^2)(1+4t^2)}\Rightarrow I=-\frac{1}{3}{\int }_0^{\infty }(\frac{1}{1+t^2}-\frac{4}{1+4t^2})dt$

$\Rightarrow I=-\frac{1}{3}{[ta{n}^{-1}t-4\frac{ta{n}^{-1}\left(2t\right)}{2}]}_0^{\infty }\therefore I=-\frac{1}{3}[(\frac{\pi }{2}-2\times \frac{\pi }{2})-(0-0\left)\right]=\frac{\pi }{6}.$