Question

${\int }_0^{\pi }\frac{xdx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$

• A. $\frac{\pi }{2ab}$
• B. $\frac{\pi }{ab}$
• C. $\frac{{\pi }^2}{2ab}$
• D. $\frac{{\pi }^2}{ab}$

Answer 1

The correct option is D $\frac{{\pi }^2}{ab}$

$I={\int }_0^{\pi }\frac{xdx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}I={\int }_0^{\pi }\frac{(\pi -x)dx}{a^2{\cos }^2(\pi -x)+b^2{\sin }^2(\pi -x)}$
Using property of definite integration ${\int }_0^{a}f\left(x\right)={\int }_0^{a}f(a-x)$
$2I=\pi {\int }_0^{\pi }\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}I=\frac{\pi }{2}{\int }_0^{\pi }\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}I=\pi {\int }_0^{\frac{\pi }{2}}\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$
Dividing numerator and denominator by ${\cos }^{2}x$
$I=\pi {\int }_0^{\frac{\pi }{2}}\frac{se{c}^{2}xdx}{a^2+b^2{\tan }^{2}x}$
let $t=b\tan xdt=b{\sec }^{2}xdx$
then
$I=\frac{\pi }{b}{\int }_0^{\infty }\frac{dt}{a^2+t^2}=\frac{\pi }{b}\frac{1}{a}{\tan }^{-1}\frac{t}{a}{]}_0^{\infty }=\frac{{\pi }^2}{ab}$