Question

Solve ${\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2}xdx}{{\cos }^{2}x+4{\sin }^{2}x}$

Answer 1

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${\int }_o^{\frac{\pi }{2}}\frac{co{s}^{2}xdx}{co{s}^{2}x+4si{n}^{2}x}$

$={\int }_o^{\frac{\pi }{2}}\frac{co{s}^{2}xdx}{co{s}^{2}x+4\left(1-{\cos }^{2}x\right)}$

$={\int }_o^{\frac{\pi }{2}}\frac{co{s}^{2}xdx}{-3co{s}^{2}x+4}$

$=-\frac{1}{3}{\int }_o^{\frac{\pi }{2}}\frac{(-3co{s}^{2}x+4)-4dx}{-3co{s}^{2}x+4}$

$=-\frac{1}{3}{\int }_o^{\frac{\pi }{2}}\left(1-\frac{4dx}{4-3co{s}^{2}x}\right)$

$=-\frac{1}{3}{\left[x\right]}_0^{\frac{\pi }{2}}+\frac{4}{3}{\int }_o^{\frac{\pi }{2}}\left(\frac{dx}{4-3co{s}^{2}x}\right)$

$=-\frac{\pi }{6}+\frac{4}{3}{\int }_o^{\frac{\pi }{2}}\left(\frac{se{c}^{2}xdx}{4{\sec }^{2}x-3}\right)$

Dividing $N^r$ and $D^r$ by $co{s}^{2}x$

$=-\frac{\pi }{6}+\frac{4}{3}{\int }_o^{\frac{\pi }{2}}\left(\frac{se{c}^{2}xdx}{4\left(1+{\tan }^{2}x\right)-3}\right)$

put $tanx=t$

$se{c}^{2}xdx=dt$

$=-\frac{\pi }{6}+\frac{4}{3}{\int }_o^{\infty }\left(\frac{dt}{4\left(1+t^2\right)-3}\right)$

$=-\frac{\pi }{6}+\frac{4}{3}{\int }_0^{\infty }\left(\frac{dt}{2t^2+1}\right)$

$=-\frac{\pi }{6}+\frac{4}{3}\times \frac{1}{2}{\left[{\tan }^{-1}\left(2t\right)\right]}_0^{\infty }=\frac{\pi }{3}$