Integral Calculus Question 109
Question: $ \int_{{}}^{{}}{\frac{\sin 2x}{1+{{\sin }^{2}}x}dx=} $
[Roorkee 1976]
Options:
A) $ \log \sin 2x+c $
B) $ \log (1+{{\sin }^{2}}x)+c $
C) $ \frac{1}{2}\log (1+{{\sin }^{2}}x)+c $
D) $ {{\tan }^{-1}}(\sin x)+c $
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ (1+{{\sin }^{2}}x)=t\Rightarrow \sin 2x,dx=dt $ Hence $ \int_{{}}^{{}}{\frac{\sin 2x}{1+{{\sin }^{2}}x}dx}=\int_{{}}^{{}}{\frac{1}{t}dt=\log (1+{{\sin }^{2}}x)+c.} $
BETA
