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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals5 Marks
Question
Evaluate :
$
\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x}{1+3 \sin ^2 x} d x
$
✓

Answer

Let
$
\begin{aligned}
I & =\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x d x}{1+3 \sin ^2 x}=\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x}{\sin ^2 x+\cos ^2 x+3 \sin ^2 x} d x \\
I & =\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x=\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x}{\cos ^2 x+4\left(1-\cos ^2 x\right)} d x \\
I & =\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x}{4-3 \cos ^2 x} d x \\
& =\frac{-1}{3} \int_0^{\frac{\pi}{2}} \frac{-3 \cos ^2 x}{4-3 \cos ^2 x} d x=\frac{-1}{3} \int_0^{\frac{\pi}{2}} \frac{4-3 \cos ^2 x-4}{4-3 \cos ^2 x} d x \\
& =\int_0^{\frac{\pi}{2}}\left(-\frac{1}{3}+\frac{4}{3\left(4-3 \cos ^2 x\right)}\right) d x \\
& =\frac{-1}{3}(x)_0^{\frac{\pi}{2}}+\frac{4}{3} \int_0^{\frac{\pi}{2}} \frac{d x}{4-3 \cos ^2 x} \\
& =-\frac{\pi}{6}+\frac{4}{3} \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{4 \sec ^2 x-3} d x
\end{aligned}
$
(On dividing by $\cos ^2 x$ in numerator and denominator)
$
=-\frac{\pi}{6}+\frac{4}{3} \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{1+4 \tan ^2 x} d x
$
Again putting $\tan x=t \Rightarrow \sec ^2 x d x=d t$,
If $x=\frac{\pi}{2}$ then $t=\tan \frac{\pi}{2}=\infty$, If $x=0$ then $t=0$
$
\begin{aligned}
I & =-\frac{\pi}{6}+\frac{4}{3} \int_0^{\infty} \frac{1}{1+4 t^2} d t \\
& =-\frac{\pi}{6}+\frac{4}{3} \times \frac{1}{2}\left(\tan ^{-1} 2 t\right)_0^{\infty} \\
I & =-\frac{\pi}{6}+\frac{2}{3} \times \frac{\pi}{2}=-\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{6} \text { Ans. }
\end{aligned}
$

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