Integral Calculus Question 459
Question: $ \int_{{}}^{{}}{{{\sin }^{3}}x\ dx} $ is equal to
[SCRA 1996]
Options:
A) $ {{\sin }^{2}}x+1 $
B) $ \sin x^{2}+x^{2}+1 $
C) $ \frac{{{\cos }^{3}}x}{3}-\cos x $
D) $ \frac{1}{4}{{\sin }^{4}}x-\frac{3}{4}{{\sin }^{2}}x $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{{{\sin }^{3}}x,dx}=\int_{{}}^{{}}{{{\sin }^{2}}x,.,\sin x,dx} $ $ =\int_{{}}^{{}}{\sin x(1-{{\cos }^{2}}x),dx} $ $ =\int_{{}}^{{}}{\sin x,dx}-\int_{{}}^{{}}{{{\cos }^{2}}x,.,\sin x,dx} $ $ =-\cos x+\frac{{{\cos }^{3}}x}{3} $ .
BETA
