Question

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

• A. $I=\frac{{\pi }^2}{4ab}$
• B. $I=\frac{{\pi }^2}{2ab}$
• C. $I=\frac{{\pi }^2}{ab}$
• D. None of these

Answer 1

The correct option is B $I=\frac{{\pi }^2}{2ab}$

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

$\Rightarrow I={\int }_0^{{}^{\pi }}\frac{\left(\pi -x\right)dx}{a^2{\cos }^2\left(\pi -x\right)+b^2{\sin }^2\left(\pi -x\right)}$

$\Rightarrow I=\pi {\int }_0^{\pi }\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}-{\int }_0^{\pi }\frac{xdx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$

$\Rightarrow I=\pi {\int }_0^{\pi }\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}-I$

$\Rightarrow 2I=\pi {\int }_0^{{}^{\pi /2}}\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$ (${\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$, if $f\left(2a-x\right)=f\left(x\right)$)

$\Rightarrow 2I=2\pi {\int }_0^{{}^{\pi /2}}\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$

$\Rightarrow I=\pi {\int }_0^{{}^{\pi /2}}\frac{{\sec }^{2}xdx}{a^2+b^2{\tan }^{2}x}$ (dividing numerator and denominator by ${\cos }^{2}x$)

putting $tanx=t$ $\Rightarrow {\sec }^{2}xdx=dt$

$\Rightarrow I=\pi {\int }_0^{{}^{\infty }}\frac{dt}{a^2+{\left(bt\right)}^2}$

$\Rightarrow I=\pi \times \frac{1}{ab}{{\left[{\tan }^{-1}\frac{bt}{a}\right]}_0}^{\infty }$

$\Rightarrow I=\frac{\pi }{ab}\left[\frac{\pi }{2}-0\right]$

$\Rightarrow I=\frac{{\pi }^2}{2ab}$