Evaluate the following Integral: ∫(tan x)/(1+m^2 tan^2 x)dx, x ∈ [0, π/2]

Question

Evaluate the following Integral

\(\int\limits_{0}^{\pi/2}\cfrac{tan\,\text x}{1+m^2tan^2\text x}d\text x\)

Answer 1

Let I = \(\int\limits_{0}^{\pi/2}\cfrac{tan\,\text x}{1+m^2tan^2\text x}d\text x\)

We have sin²x + cos²x = 1

Put sin²x = t

⇒ 2 sin x cos x dx = dt (Differentiating both sides)

⇒ sin x cos x dx = \(\cfrac12\)dt

When x = 0, t = sin²0 = 0

When x = \(\cfrac{\pi}2\), t = sin²\(\cfrac{\pi}2\) = 1

So, the new limits are 0 and 1.

Substituting this in the original integral,