Search any question & find its solution
Question:
Answered & Verified by Expert
By using the properties of definite integrals, evaluate the integrals
$\int_0^{\frac{\pi}{2}} \frac{\sin ^{3 / 2} x d x}{\sin ^{3 / 2} x+\cos ^{3 / 2} d x}$
$\int_0^{\frac{\pi}{2}} \frac{\sin ^{3 / 2} x d x}{\sin ^{3 / 2} x+\cos ^{3 / 2} d x}$
Solution:
2470 Upvotes
Verified Answer
Let $\mathrm{I}=\int_0^{\pi / 2} \frac{\sin ^{3 / 2} x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x} d x \quad \ldots(i)$
Also $\mathrm{I}=\int_0^{\pi / 2} \frac{\sin ^{3 / 2}\left(\frac{\pi}{2}-x\right)}{\sin ^{3 / 2}\left(\frac{\pi}{2}-x\right)+\cos ^{3 / 2}\left(\frac{\pi}{2}-x\right)} d x$
$=\int_0^{\pi / 2} \frac{\cos ^{3 / 2} x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x} d x \quad \ldots(ii)$
Adding (i) and (ii), we have
$2 \mathrm{I}=\int_0^{\pi / 2} 1 d x=[x]_0^{\pi / 2}=\frac{\pi}{2}-0=\frac{\pi}{2} \therefore \mathrm{I}=\frac{\pi}{4}$
Also $\mathrm{I}=\int_0^{\pi / 2} \frac{\sin ^{3 / 2}\left(\frac{\pi}{2}-x\right)}{\sin ^{3 / 2}\left(\frac{\pi}{2}-x\right)+\cos ^{3 / 2}\left(\frac{\pi}{2}-x\right)} d x$
$=\int_0^{\pi / 2} \frac{\cos ^{3 / 2} x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x} d x \quad \ldots(ii)$
Adding (i) and (ii), we have
$2 \mathrm{I}=\int_0^{\pi / 2} 1 d x=[x]_0^{\pi / 2}=\frac{\pi}{2}-0=\frac{\pi}{2} \therefore \mathrm{I}=\frac{\pi}{4}$
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.