Discussion Forum

$\int_{0}^{\pi} \frac{xdx}{4\cos^{2}x+9\sin^{2}x}=$

Answer:

Explanation:

We have,

 $I=\int_{0}^{\pi} \frac{xdx}{4 \cos^{2}x+9 \sin^{2}x}$

 $\Rightarrow$       $I=\int_{0}^{\pi} \frac{(\pi-x)dx}{4 \cos^{2}(\pi-x)+9 \sin^{2}(\pi-x)}$  

$\Rightarrow$   $I=\int_{0}^{\pi} \frac{(\pi-x)dx}{4 \cos^{2}x+9 \sin^{2}x}$

$\Rightarrow$     $2I=\int_{0}^{\pi} \frac{\pi dx}{4 \cos^{2}x+9 \sin^{2}x}$

$\Rightarrow$     $2I=\int_{0}^{\pi} \frac{\pi \sec^{2}x dx}{4 +9 \tan^{2}x}$

$\Rightarrow$    $2I=\int_{0}^{\pi/2} \frac{2\pi \sec^{2} x dx}{4 +9 \tan^{2}x}$

            $\left[\because \int_{0}^{2a} f(x) dx-2\int_{0}^{a} f(x) dx\Rightarrow f(2a-x)=f(x)\right]$

$\Rightarrow$      $I= \frac{\pi}{9}\int_{0}^{\pi/2} \frac{\sec^{2} x dx}{\frac{4}{9}+\tan^{2}x }$

 Put   $\tan x=t \Rightarrow  \sec^{2} x dx=dt$

       $x=0, t=0, x=\frac{\pi}{2},t=\infty$

  $\Rightarrow$    $I= \frac{\pi}{9} \int_{0}^{\infty} \frac{dt}{\left(\frac{2}{3}\right)^{2}+t^{2}}$

$\Rightarrow$    $I= \frac{\pi}{9} \times \frac{3}{2}\left[ \tan ^{-1} \frac{3t}{2}\right]_{0}^{\infty}$

$\Rightarrow$     $I= \frac{\pi}{9} \times \frac{3}{2}\times \frac{\pi}{2}= \frac{\pi^{2}}{12}$