Integral Calculus Question 57
Question: $ \int_{{}}^{{}}{2x{{\cos }^{3}}x^{2}\sin x^{2}dx=} $
Options:
A) $ -\frac{1}{4}{{\cos }^{4}}x^{2}+c $
B) $ \frac{1}{4}{{\cos }^{4}}x^{2}+c $
C) $ {{\cos }^{4}}x^{2}+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Put $ t=\cos x^{2}\Rightarrow dt=-2x\sin x^{2}dx, $ then $ \int_{{}}^{{}}{2x{{\cos }^{3}}x^{2}\sin x^{2}dx}=-\int_{{}}^{{}}{t^{3}dt}=-\frac{t^{4}}{4}+c $ $ =-\frac{1}{4}{{\cos }^{4}}x^{2}+c. $
BETA
