Evaluate the following Integral: ∫cos^2x/(1+3sin^2x)dx, x ∈ [0,π/2]

Question

Evaluate the following Integral:

\(\int\limits_{0}^{\pi/2}\cfrac{cos^2\text x}{1+3sin^2\text x}d\text x\)

Answer 1

Let I =\(\int\limits_{0}^{\pi/2}\cfrac{cos^2\text x}{1+3sin^2\text x}d\text x\)

Dividing numerator and denominator with cos 2x, we have I = \(\int\limits_{0}^{\pi/2}\cfrac{1}{sec^2\text x+3tan^2\text x}d\text x\)

⇒\(\int\limits_{0}^{\pi/2}\cfrac{1}{1+4tan^2\text x}d\text x\) [\(\because\) sec² x = 1 + tan²x]

Put tan x = t

⇒ sec²x dx = dt (Differentiating both sides)

⇒ dx = \(\cfrac{dt}{1+tan^2\text x}\) = \(\cfrac{dt}{1+t^2}\)

When x = 0, t = tan 0 = 0

When x = \(\cfrac{\pi}2\), t = tan\(\cfrac{\pi}2\)= ∞

So, the new limits are 0 and ∞.

Substituting this in the original integral,

Multiplying numerator and denominator with 3, we have

Now, we can write

Substituting this in the original integral,