Question

Evaluate : $\int\limits^{\pi/2}_0 { \frac{1}{a^{2}\sin^{2}x+b^{2}\cos^{2}x}} \, dx$

Answer 1

we have to evaluate $\int\limits^{\pi/2}_0 { \frac{1}{a^{2}\sin^{2}x+b^{2}\cos^{2}x}}\,dx$

numerator and denominator are divided by cos²x .

= $\int\limits^{\pi/2}_0{\frac{sec^2x}{a^{2}tan^{2}x+b^{2}}}\,dx$

now, let tanx = t

differentiating both sides,

sec²x dx = dt

lower limit, t = tan(0) = 0

upper limit , t = tan(π/2) = ∞

so, $\int\limits^{\pi/2}_0 { \frac{1}{a^{2}\sin^{2}x+b^{2}\cos^{2}x}} \, dx$ converted into $\int\limits^{\infty}_0{\frac{1}{a^2t^2+b^2}}\,dt$

= $\int\limits^{\infty}_0{\frac{1}{(at)^2+b^2}}\,dt$

= $\frac{1}{ab}\left[tan^{-1}\frac{at}{b}\right]^{\infty}_0$

= $\frac{1}{ab}\left[\frac{\pi}{2}-0\right]$

= $\frac{\pi}{2ab}$

Answer 2

Answer:

$\int\limits^{\pi/2}_0 { \frac{1}{a^{2}\sin^{2}x+b^{2}\cos^{2}x}}\ dx=\frac{\pi}{2ab}$

Step-by-step explanation:

In this question,

We need to find out the value of,

$\int\limits^{\pi/2}_0 { \frac{1}{a^{2}\sin^{2}x+b^{2}\cos^{2}x}}\ dx$

Now, divide the Numerator and Denominator by cos²x we get,

$= \int\limits^{\pi/2}_0{\frac{sec^2x}{a^{2}tan^{2}x+b^{2}}}\,dx$

Now, let us say that,

tan x = t

So, on differentiating, we get,

dt = (sec²x)dx

Also,

at, x = 0,

t = tan(0) = 0

and,

At, x = π/2

t = tan(π/2) = ∞

So, on putting this in the above equation we get,

$\int\limits^{\infty}_0{\frac{1}{a^2t^2+b^2}}\,dt$

Therefore, on solving this further we get,

$= \int\limits^{\infty}_0{\frac{1}{(at)^2+b^2}}\,dt\\=\frac{1}{ab}\left[tan^{-1}\frac{at}{b}\right]^{\infty}_0\\= \frac{1}{ab}\left[\frac{\pi}{2}-0\right]\\=\frac{\pi}{2ab}$

Therefore, on integrating the given equation we get,

$\frac{\pi}{2ab}$