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Definite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_0^{\pi / 2} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x=$
  • A
    $0$
  • B
    $\frac{\pi}{2}$
  • ✓
    $\frac{\pi}{4}$
  • D
    $\frac{\pi}{3}$
✓

Answer

Correct option: C.
$\frac{\pi}{4}$
(C)
Let $I =\int_0^{\pi / 2} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$ ...(i)
$\therefore \quad I =\int_0^{\pi / 2} \frac{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x$
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$\therefore \quad I=\int_0^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x$ ...(ii)
Adding (i) and (ii), we get
$2 I =\int_0^{\pi / 2} d x=\frac{\pi}{2} \Rightarrow I =\frac{\pi}{4}$
Alternate Method:
$\int_0^{\frac{\pi}{2}} \frac{\sin ^{ n } x}{\sin ^{ n } x+\cos ^{ n } x} d x=\int_0^{\frac{\pi}{2}} \frac{\cos ^{ n } x}{\sin ^{ n } x+\cos ^{ n } x} d x=\frac{\pi}{4}$
$\int_0^{\pi / 2} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x=\frac{\pi}{4}$

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