Definite Integration Question 246
Question: $ \int_0^{\pi /2}{\frac{{{\sin }^{3/2}}xdx}{{{\cos }^{3/2}}x+{{\sin }^{3/2}}x}}= $
[Roorkee 1989; BIT Ranchi 1989]
Options:
A) 0
B) $ \pi $
C) $ \pi /2 $
D) $ \pi /4 $
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ I=\int_0^{\pi /2}{\frac{{{\sin }^{3/2}}xdx}{{{\cos }^{3/2}}x+{{\sin }^{3/2}}x}} $ ……(i) = $ \int_0^{\pi /2}{\frac{{{\sin }^{3/2}}( \frac{\pi }{2}-x )}{{{\cos }^{3/2}}( \frac{\pi }{2}-x )+{{\sin }^{3/2}}( \frac{\pi }{2}-x )}dx} $
= $ \int_0^{\pi /2}{\frac{{{\cos }^{3/2}}xdx}{{{\sin }^{3/2}}x+{{\cos }^{3/2}}x}} $ ……(ii)
Adding (i) and (ii), we get $ I=\frac{1}{2}\int_0^{\pi /2}{1dx=\frac{1}{2}[x]_0^{\pi /2}=\frac{\pi }{4}} $ .
BETA
