Definite Integration Question 248
Question: $ \int _{-\pi /2}^{\pi /2}{{{\sin }^{2}}x{{\cos }^{2}}x(\sin x+\cos x)dx=} $
[EAMCET 1992]
Options:
A) $ \frac{2}{15} $
B) $ \frac{4}{15} $
C) $ \frac{6}{15} $
D) $ \frac{8}{15} $
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Answer:
Correct Answer: B
Solution:
$ \int _{-\pi /2}^{\pi /2}{{{\sin }^{2}}x{{\cos }^{2}}x(\sin x+\cos x)dx} $
= $ \int _{-\pi /2}^{\pi /2}{{{\sin }^{3}}x{{\cos }^{2}}xdx+\int _{-\pi /2}^{\pi /2}{{{\sin }^{2}}x{{\cos }^{3}}xdx}} $
$ =0+2\int_0^{\pi /2}{{{\sin }^{2}}x{{\cos }^{3}}xdx} $
$ =0+2\times \frac{2}{15}=\frac{4}{15} $ .
BETA
