Integral Calculus Question 394
Question: $ \int_{{}}^{{}}{\frac{\sin 2x}{a^{2}+b^{2}{{\sin }^{2}}x}}\ dx= $
[Roorkee 1977]
Options:
A) $ \frac{1}{b^{2}}\log (a^{2}+b^{2}{{\sin }^{2}}x)+c $
B) $ \frac{1}{b}\log (a^{2}+b^{2}{{\sin }^{2}}x)+c $
C) $ \log (a^{2}+b^{2}{{\sin }^{2}}x)+c $
D) $ b^{2}\log (a^{2}+b^{2}{{\sin }^{2}}x)+c $
Show Answer
Answer:
Correct Answer: A
Solution:
Put $ a^{2}+b^{2}{{\sin }^{2}}x=t\Rightarrow b^{2}\sin 2x,dx=dt, $ then $ \int_{{}}^{{}}{\frac{\sin 2x}{a^{2}+b^{2}{{\sin }^{2}}x},dx=\frac{1}{b^{2}}\int_{{}}^{{}}{\frac{dt}{t}=\frac{1}{b^{2}}\log t+c}} $ $ =\frac{1}{b^{2}}\log (a^{2}+b^{2}{{\sin }^{2}}x)+c. $
BETA
