Integral Calculus
86259
Evaluate : \(\int_{0}^{\pi / 2} \frac{1}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x} d x\)
1 \(\frac{\pi \mathrm{a}}{4 \mathrm{~b}}\)
2 \(\frac{\pi \mathrm{a}}{2 \mathrm{~b}}\)
3 \(\frac{\pi \mathrm{b}}{4 \mathrm{a}}\)
4 \(\frac{\pi}{2 \mathrm{ab}}\)
Explanation:
(D) : Let,
\(I=\int_{0}^{\pi / 2} \frac{1}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x} d x\)
Dividing numerator and denominator by \(\cos ^{2} x\) we get
\(I=\int_{0}^{\pi / 2} \frac{\sec ^{2} x d x}{a^{2} \tan ^{2} x+b^{2}}\)
Put, \(\tan x=t \Rightarrow \sec ^{2} x d x=d t\)
When, \(x=0 \Rightarrow t=0\) and when \(x=\frac{\pi}{2} \Rightarrow t=\infty\)
\(\Rightarrow \mathrm{I}=\int_{0}^{\infty} \frac{\mathrm{dt}}{\mathrm{a}^{2} \mathrm{t}^{2}+\mathrm{b}^{2}}=\frac{1}{\mathrm{a}^{2}} \int_{0}^{\infty} \frac{\mathrm{dt}}{\mathrm{t}^{2}+\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{2}}\)
\(=\frac{1}{\mathrm{a}^{2}} \times \frac{1}{\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}\left[\tan ^{-1} \frac{\mathrm{t}}{(\mathrm{b} / \mathrm{a})}\right]_{0}^{\infty}=\frac{1}{\mathrm{ab}} \times \frac{\pi}{2}=\frac{\pi}{2 \mathrm{ab}}\)
