Integral Calculus Question 462
Question: $ \int_{{}}^{{}}{{{\sin }^{2}}x\cos x\ dx} $ is equal to
[SCRA 1996]
Options:
A) $ \frac{{{\cos }^{2}}x}{2}+c $
B) $ \frac{{{\sin }^{2}}x}{3}+c $
C) $ \frac{{{\sin }^{3}}x}{3}+c $
D) $ -\frac{{{\cos }^{2}}x}{2}+c $
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Answer:
Correct Answer: C
Solution:
$ I=\int_{{}}^{{}}{{{\sin }^{2}}x,.,\cos x,dx} $ . Put $ \sin x=t\Rightarrow \cos x,dx=dt $
$ \therefore ,I=\int_{{}}^{{}}{t^{2}dt}=\frac{t^{3}}{3}+c=\frac{{{\sin }^{3}}x}{3}+c $ .
BETA
