Integral Calculus Question 461
Question: $ \int_{{}}^{{}}{\sin \sqrt{x}}\ dx= $
[Roorkee 1977]
Options:
A) $ 2[\sin \sqrt{x}-\cos \sqrt{x}]+c $
B) $ 2[\sin \sqrt{x}-\sqrt{x}\cos \sqrt{x}]+c $
C) $ 2[\sin \sqrt{x}+\cos \sqrt{x}]+c $
D) $ 2[\sin \sqrt{x}+\sqrt{x}\cos \sqrt{x}]+c $
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ \sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}},dx=dt\Rightarrow dx=2t,dt, $ then
$ \int_{{}}^{{}}{\sin \sqrt{x},dx}=2\int_{{}}^{{}}{t\sin t,dt}=2(-t\cos t+\sin t)+c $
$ =2(\sin \sqrt{x}-\sqrt{x}\cos \sqrt{x})+c. $
BETA
