Question

Evaluate ${\int }_0^{\frac{\pi }{4}}\frac{dx}{a^2{\cos }^{2}x-b^2{\sin }^{2}x}$

Answer 1

${\int }_0^{\frac{\pi }{4}}\frac{dx}{a^2{\cos }^{2}x-b^2{\sin }^{2}x}$

$={\int }_0^{\frac{\pi }{4}}\frac{\frac{1}{{\cos }^{2}x}dx}{a^2-b^2\frac{{\sin }^{2}x}{{\cos }^{2}x}}$

$={\int }_0^{\frac{\pi }{4}}\frac{{\sec }^{2}xdx}{a^2-b^2{\tan }^{2}x}$

$=\frac{1}{a^2}{\int }_0^{\frac{\pi }{4}}\frac{{\sec }^{2}xdx}{1-\frac{b^2}{a^2}{\tan }^{2}x}$

Let $t=\frac{b}{a}\tan x\Rightarrow dt=\frac{b}{a}{\sec }^{2}xdx$

When $x=0\Rightarrow t=0$

When $x=\frac{\pi }{4}\Rightarrow t=\frac{b}{a}$

$=\frac{1}{a^2}\times \frac{a}{b}{\int }_0^{\frac{b}{a}}\frac{dt}{1-t^2}$

$=\frac{1}{ab}{\int }_0^{\frac{b}{a}}\frac{dt}{1-t^2}$

We know that $\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\log \left|\frac{a+x}{a-x}\right|+c$

$=\frac{1}{ab}\times \frac{1}{2}{\left[\log \left|\frac{1+t}{1-t}\right|\right]}_0^{\frac{b}{a}}$

$=\frac{1}{2ab}[\log \left|\frac{1+\frac{b}{a}}{1-\frac{b}{a}}\right|-\log \left|\frac{1+0}{1-0}\right|]$

$=\frac{1}{2ab}\log \left|\frac{a+b}{a-b}\right|$