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Mathematics Question on Definite Integral

If I=∫0π/4(tan⁡x+cot⁡x) dxI = \int_{0}^{\pi/4} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) \, dx, then value of I is

A

π2\frac{\pi}{2}

B

π22\frac{\pi}{2\sqrt{2}}

C

π3\frac{\pi}{\sqrt{3}}

D

π2\frac{\pi}{\sqrt{2}}

Answer

π2\frac{\pi}{\sqrt{2}}

Explanation

Solution

We have,
I=∫0π4(tan⁡x+cot⁡x) dxI = \int_{0}^{\frac{\pi}{4}} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) \, dx
⇒I=∫0π4(sin⁡xcos⁡x+cos⁡xsin⁡x) dx\Rightarrow I = \int_{0}^{\frac{\pi}{4}} \left( \sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}} \right) \, dx
⇒I=∫0π4sin⁡x+cos⁡xsin⁡xcos⁡x dx\Rightarrow I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx
Put sin⁡x−cos⁡x=t\sin x - \cos x = t
⇒(cos⁡x+sin⁡x) dx=dt\Rightarrow (\cos x + \sin x) \, dx = dt
x=0,t=−1x = 0, t = -1
and x=π4⇒t=0x = \frac{\pi}{4} \Rightarrow t = 0
⇒I=∫−102 dt1−t2\Rightarrow I = \int_{-1}^{0} \frac{\sqrt{2} \, dt}{\sqrt{1 - t^{2}}}
⇒I=2[sin⁡−1t]−10\Rightarrow I = \sqrt{2} \left[ \sin^{-1} t \right]_{-1}^{0}
⇒I=2[sin⁡−10−sin⁡−1(−1)]\Rightarrow I = \sqrt{2} \left[ \sin^{-1} 0 - \sin^{-1}(-1) \right]
⇒I=2[0+π2]\Rightarrow I = \sqrt{2} \left[ 0 + \frac{\pi}{2} \right]
⇒I=π2\Rightarrow I = \frac{\pi}{\sqrt{2}}