Question

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

• A. $\frac{\pi^2}{12}$
• B. $\frac{\pi^2}{4}$
• C. $\frac{\pi^2}{6}$
• D. $\frac{\pi^2}{3}$

Answer 1

Answer: (A)

We have,
$ I =\int_\limits{0}^{\pi} \frac{x d x}{4 \cos ^{2} x+9 \sin ^{2} x} $
$ \Rightarrow I=\int\limits_{0}^{\pi} \frac{(\pi-x) d x}{4 \cos ^{2}(\pi-x)+9 \sin ^{2}(\pi-x)}$
$ \Rightarrow I=\int_\limits{0}^{\pi} \frac{(\pi-x) d x}{4 \cos ^{2} x+9 \sin ^{2} x} $
$ \Rightarrow 2 I=\int_\limits{0}^{\pi} \frac{\pi d x}{4 \cos ^{2} x+9 \sin ^{2} x} $
$\Rightarrow 2 I=\int_\limits{0}^{\pi} \frac{\pi \sec ^{2} x d x}{4+9 \tan ^{2} x} $
$ \Rightarrow 2 I=\int_\limits {0}^{\pi / 2} \frac{2 \pi \sec ^{2} x d x}{4+9 \tan ^{2} x} $
$[\because \int_\limits{0}^{2 a} f(x) d x-2 \int_\limits{0}^{a} f(x) d x \Rightarrow f(2 a-x)=f(x)]$
$\Rightarrow I=\frac{\pi}{9} \int_\limits{0}^{\pi / 2} \frac{\sec ^{2} x d x}{\frac{4}{9}+\tan ^{2} x}$
Put $\tan x=t \Rightarrow \sec ^{2} x d x=d t$
$x=0, t=0, x=\frac{\pi}{2}, t=\infty$
$\Rightarrow I=\frac{\pi}{9} \int_\limits{0}^{\infty} \frac{d t}{\left(\frac{2}{3}\right)^{2}+t^{2}}$
$\Rightarrow I=\frac{\pi}{9} \times \frac{3}{2}\left[\tan ^{-1} \frac{3 t}{2}\right]_{0}^{\infty}$
$\Rightarrow I=\frac{\pi}{9} \times \frac{3}{2} \times \frac{\pi}{2}=\frac{\pi^{2}}{12}$