Integral Calculus Question 447
Question: $ \int_{{}}^{{}}{x\cos x^{2}\ dx} $ is equal to
[MP PET 1999; Pb. CET 2000]
Options:
A) $ -\frac{1}{2}{{\sin }^{2}}x+c $
B) $ \frac{1}{2}{{\sin }^{2}}x+c $
C) $ -\frac{1}{2}\sin x^{2}+c $
D) $ \frac{1}{2}\sin x^{2}+c $
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Answer:
Correct Answer: D
Solution:
Put $ x^{2}=t\Rightarrow dt=2x,dx $ \ Given integral $ =\frac{1}{2}\int_{{}}^{{}}{\cos t,dt}=\frac{1}{2}\sin t=\frac{1}{2}\sin x^{2}+c $ .
BETA
