Search any question & find its solution
Question:
Answered & Verified by Expert
$\int_0^{\pi / 2} \frac{\sin ^{3 / 2} x d x}{\cos ^{3 / 2} x+\sin ^{3 / 2} x}=$
Options:
Solution:
1028 Upvotes
Verified Answer
The correct answer is:
$\pi / 4$
Let $I=\int_0^{\pi / 2} \frac{\sin ^{3 / 2} x d x}{\cos ^{3 / 2} x+\sin ^{3 / 2} x}$ ...(i)
$=\int_0^{\pi / 2} \frac{\sin ^{3 / 2}\left(\frac{\pi}{2}-x\right)}{\cos ^{3 / 2}\left(\frac{\pi}{2}-x\right)+\sin ^{3 / 2}\left(\frac{\pi}{2}-x\right)} d x$
$=\int_0^{\pi / 2} \frac{\cos ^{3 / 2} x d x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x}$ ...(ii)
Adding (i) and (ii), we get $I=\frac{1}{2} \int_0^{\pi / 2} 1 d x=\frac{1}{2}[x]_0^{\pi / 2}=\frac{\pi}{4}$
$=\int_0^{\pi / 2} \frac{\sin ^{3 / 2}\left(\frac{\pi}{2}-x\right)}{\cos ^{3 / 2}\left(\frac{\pi}{2}-x\right)+\sin ^{3 / 2}\left(\frac{\pi}{2}-x\right)} d x$
$=\int_0^{\pi / 2} \frac{\cos ^{3 / 2} x d x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x}$ ...(ii)
Adding (i) and (ii), we get $I=\frac{1}{2} \int_0^{\pi / 2} 1 d x=\frac{1}{2}[x]_0^{\pi / 2}=\frac{\pi}{4}$
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.