Integral Calculus Question 396
Question: $ \int_{{}}^{{}}{\cos \sqrt{x}\ dx=} $
[BIT Ranchi 1990; IIT 1977; RPET 1999]
Options:
A) $ 2[\sqrt{x}\sin \sqrt{x}+\cos \sqrt{x}]+c $
B) $ 2[\sqrt{x}\sin \sqrt{x}-\cos \sqrt{x}]+c $
C) $ 2[\cos \sqrt{x}-\sqrt{x}\sin \sqrt{x}]+c $
D) $ -2[\sqrt{x}\sin \sqrt{x}+\cos \sqrt{x}]+c $
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Answer:
Correct Answer: A
Solution:
Put $ \sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}},dx=dt\Rightarrow dx=2t,dt, $ then it reduces to $ \int_{{}}^{{}}{2t,.\cos t,dt}=2[ t,.,\sin t-\int_{{}}^{{}}{\sin t,dt} ] $ $ =2t\sin t+2\cos t $ $ =2[\sqrt{x}\sin \sqrt{x}+\cos \sqrt{x}]+c $ .
BETA
