Question

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

Answer 1

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

$=\frac{1}{2}\int \frac{dx}{\cos x+2\sin x}$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{dx}{\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)+2\left(\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)}$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{(1+{\tan }^2\frac{x}{2})dx}{1-{\tan }^2\frac{x}{2}+4\tan \frac{x}{2}}$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{{\sec }^2\frac{x}{2}dx}{1-{\tan }^2\frac{x}{2}+4\tan \frac{x}{2}}$

Let $t=\tan \frac{x}{2}\Rightarrow dt=\frac{1}{2}{\sec }^2\frac{x}{2}dx$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{2dt}{1-t^2+4t}$

$=-{\int }_0^{\frac{\pi }{2}}\frac{dt}{t^2-4t-1}$

$=-{\int }_0^{\frac{\pi }{2}}\frac{dt}{t^2-2\times 2t+4-4-1}$

$=-{\int }_0^{\frac{\pi }{2}}\frac{dt}{{(t-2)}^2-{\left(\sqrt{5}\right)}^2}$

$={\int }_0^{\frac{\pi }{2}}\frac{dt}{{\left(\sqrt{5}\right)}^2-{(t-2)}^2}$

We know that $\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\ln \frac{a+x}{a-x}+c$

$={\left[\frac{1}{2\times \sqrt{5}}\ln \frac{\sqrt{5}+t-2}{\sqrt{5}-t+2}\right]}_0^{\frac{\pi }{2}}$

$={\left[\frac{1}{2\times \sqrt{5}}\ln \frac{\sqrt{5}+\tan \frac{x}{2}-2}{\sqrt{5}-\tan \frac{x}{2}+2}\right]}_0^{\frac{\pi }{2}}$ where $t=\tan \frac{x}{2}$

$=\frac{1}{2\sqrt{5}}[\ln \frac{\sqrt{5}+\tan \frac{\pi }{4}-2}{\sqrt{5}-\tan \frac{\pi }{4}+2}-\ln \frac{\sqrt{5}+\tan 0-2}{\sqrt{5}-\tan 0+2}]$

$=\frac{1}{2\sqrt{5}}[\ln \frac{\sqrt{5}+1-2}{\sqrt{5}-1+2}-\ln \frac{\sqrt{5}+0-2}{\sqrt{5}-0+2}]$

$=\frac{1}{2\sqrt{5}}[\ln \frac{\sqrt{5}-1}{\sqrt{5}+1}-\ln \frac{\sqrt{5}-2}{\sqrt{5}+2}]$