Search any question & find its solution
Question:
Answered & Verified by Expert
By using the properties of definite integrals, evaluate the integrals
$\int_0^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$
$\int_0^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$
Solution:
1967 Upvotes
Verified Answer
Let $\mathrm{I}=\int_0^{\pi / 2} \frac{\sqrt{\sin \mathrm{x}}}{\sqrt{\sin \mathrm{x}}+\sqrt{\cos \mathrm{x}}} \mathrm{dx} \quad \ldots(i)$
$I=\int_0^{\pi / 2} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x$
$I=\int_0^{\pi / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \quad \ldots(ii)$
Adding eqs. (i) \& (ii), we get
$\begin{aligned}
&2 \mathrm{I}=\int_0^{\pi / 2} 1 \cdot \mathrm{dx}=[\mathrm{x}]_0^{\pi / 2}=\frac{\pi}{2} \\
&\therefore \quad \mathrm{I}=\frac{\pi}{4}
\end{aligned}$
$I=\int_0^{\pi / 2} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x$
$I=\int_0^{\pi / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \quad \ldots(ii)$
Adding eqs. (i) \& (ii), we get
$\begin{aligned}
&2 \mathrm{I}=\int_0^{\pi / 2} 1 \cdot \mathrm{dx}=[\mathrm{x}]_0^{\pi / 2}=\frac{\pi}{2} \\
&\therefore \quad \mathrm{I}=\frac{\pi}{4}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.