Integral Calculus Question 82
Question: $ \int \frac{2\sin x}{(3+sin2x)}dx $ is equal to
Options:
A) $ \frac{1}{2}\ln | \frac{2+\sin x-\cos x}{2-\sin x+\cos x} |-\frac{1}{\sqrt{2}}{{\tan }^{-1}}( \frac{\sin x+\cos x}{\sqrt{2}} )+c $
B) $ \frac{1}{2}\ln | \frac{2+\sin x-\cos x}{2-\sin x+\cos x} |-\frac{1}{2\sqrt{2}}{{\tan }^{-1}}( \frac{\sin x+\cos x}{\sqrt{2}} )+c $
C) $ \frac{1}{4}\ln | \frac{2+\sin x-\cos x}{2-\sin x+\cos x} |-\frac{1}{\sqrt{2}}{{\tan }^{-1}}( \frac{\sin x+\cos x}{\sqrt{2}} )+c $
D) none of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ I=\int{\frac{2\sin x}{(3+sin2x)}dx} $ $ =\int{\frac{\sin x+\cos x+\sin x-\cos x}{(3+sin2x)}dx} $ $ =\int{\frac{\sin x+\cos x}{3+\underset{\begin{smallmatrix} \downarrow \\ I_1 \end{smallmatrix}}{\mathop{\sin }},2x}}dx-\int{\frac{-\sin x+\cos x}{(3+\underset{\begin{smallmatrix} \downarrow \\ I_2 \end{smallmatrix}}{\mathop{\sin }},2x)}dx} $ Putting $ t_1=\sin x-\cos x $ in $ I_1 $ and $ t_2=\sin x+\cos x $ in $ I_2 $ , we get $ I=\int{\frac{dt_1}{[3+(1-t^2_1)]}-\int{\frac{dt_2}{[3+(t_2^{2}-1)]}}} $ = $ \int{\frac{dt}{4-t_1^{2}}-\int{\frac{dt_2}{2+t_2^{2}}}} $ $ =\frac{1}{4}\ln | \frac{2+t_1}{2-t_1} |-\frac{1}{\sqrt{2}}{{\tan }^{-1}}( \frac{t_2}{\sqrt{2}} )+C $ $ =\frac{1}{4}\ln | \frac{2+\sin x-\cos x}{2-\sin x+\cos x} |-\frac{1}{\sqrt{2}}{{\tan }^{-1}}( \frac{\sin x+\cos x}{\sqrt{2}} )+C $
BETA
