Integral Calculus Question 124
Question: $ \int_{{}}^{{}}{{{\sin }^{3}}x\ .\ \cos x\ dx=} $
[SCRA 1996]
Options:
A) $ \frac{{{\sin }^{4}}x{{\cos }^{2}}x}{8}+c $
B) $ \frac{{{\sin }^{4}}x}{4}+c $
C) $ \frac{{{\sin }^{2}}x}{2}+c $
D) $ 4{{\sin }^{4}}x+c $
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Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{{{\sin }^{3}}x,.,\cos x,dx} $ . Put $ \sin x=t, $ then $ \cos x,dx=dt $ ; $ \int_{{}}^{{}}{t^{3}dt}=\frac{t^{4}}{4}=\frac{{{\sin }^{4}}x}{4}+c $ .
BETA
