Integral Calculus Question 416
Question: $ \int_{{}}^{{}}{\cos x\sqrt{4-{{\sin }^{2}}x}}\ dx= $
Options:
A) $ \frac{1}{2}\sin x\sqrt{4-{{\sin }^{2}}x}-2{{\sin }^{-1}}( \frac{1}{2}\sin x )+c $
B) $ \frac{1}{2}\sin x\sqrt{4-{{\sin }^{2}}x}+2{{\sin }^{-1}}( \frac{1}{2}\sin x )+c $
C) $ \frac{1}{2}\sin x\sqrt{4-{{\sin }^{2}}x}+{{\sin }^{-1}}( \frac{1}{2}\sin x )+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Putting $ \sin x=t\Rightarrow \cos x,dx=dt, $ we get $ \int_{{}}^{{}}{\cos x\sqrt{4-{{\sin }^{2}}x,}dx}=\int_{{}}^{{}}{\sqrt{4-t^{2}}dt=\int_{{}}^{{}}{\sqrt{{{(2)}^{2}}-t^{2}}dt}} $ $ =\frac{t}{2}\sqrt{4-t^{2}}+\frac{4}{2}{{\sin }^{-1}}\frac{t}{2}+c $ $ =\frac{1}{2}\sin x\sqrt{4-{{\sin }^{2}}x}+2{{\sin }^{-1}}( \frac{1}{2}\sin x )+c. $
BETA
