Question

`∫_0^(π/2) (sqrttan x + sqrtcot x)dx` = ______.

Options

• `π/sqrt2`

• `πsqrt2`

• `π/2`

• `sqrt2/π`

Answer 1

`∫_0^(π/2) (sqrttan x + sqrtcot x)dx` = `bbunderline(πsqrt2)`.

Explanation:

Let I = `∫_0^(π/2) (sqrttan x + sqrtcot x)dx`

`∫_0^(π/2) (sin x + cos x)/sqrt(sin x cos x)`dx

`∫_0^(π/2) (sqrt2(sin x + cos x))/sqrt(sin2x)`dx

= `∫_0^(π/2) (sqrt2(sin x + cos x))/sqrt(I − sin x − cos x)^2`dx

Put, sin x - cos x = t

⇒ (cos x + sin x) dx = dt

`∫_(−1)^(1) sqrt2/sqrt(1 − t^2)dt = sqrt2 sin^(−1) t|_(−1)^(1)`

`sqrt2[π/2 + π/2] = πsqrt2`