Question

$\underset{0}{\overset{\pi /2}{\int }}\left(\frac{1+\sin 3x}{1+2\sin x}\right)dx$ is equal to

• A. $\underset{0}{\overset{\pi /2}{\int }}\left(\frac{1-\cos 3x}{1+2\cos x}\right)dx$
• B. $\underset{0}{\overset{\pi /2}{\int }}\left(\frac{\cos 3x+1}{2\cos x-1}\right)dx$
• C. $\frac{\pi }{2}$
• D. $1$

Answer 1

Answer: (A), (B), (D)

$1$

Let $I=\underset{0}{\overset{\pi /2}{\int }}\left(\frac{1+\sin 3x}{1+2\sin x}\right)dx$
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}\left(\frac{1+\sin 3(\frac{\pi }{2}-x)}{1+2\sin (\frac{\pi }{2}-x)}\right)dx$
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}\left(\frac{1-\cos 3x}{1+2\cos x}\right)dx$

Now, $I=-\underset{0}{\overset{\pi /2}{\int }}\left(\frac{\cos 3x-1}{1+2\cos x}\right)dx$
$\Rightarrow I=-\underset{0}{\overset{\pi /2}{\int }}\left(\frac{4{\cos }^{3}x-3\cos x-1}{1+2\cos x}\right)dx$
$\Rightarrow I=-\underset{0}{\overset{\pi /2}{\int }}\left(\frac{4{\cos }^{3}x+2{\cos }^{2}x-2{\cos }^{2}x-\cos x-2\cos x-1}{1+2\cos x}\right)dx$
$\Rightarrow I=-\underset{0}{\overset{\pi /2}{\int }}\frac{(1+2\cos x)(2{\cos }^{2}x-\cos x-1)}{1+2\cos x}dx$
$\Rightarrow I=-\underset{0}{\overset{\pi /2}{\int }}(\cos 2x-\cos x)dx$
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}(\cos x-\cos 2x)dx$
$=1-0=1$


Similarly, $I_1=\underset{0}{\overset{\pi /2}{\int }}\left(\frac{\cos 3x+1}{2\cos x-1}\right)dx$
$\Rightarrow I_1=\underset{0}{\overset{\pi /2}{\int }}\frac{(2\cos x-1)(2{\cos }^{2}x+\cos x-1)}{2\cos x-1}dx$
$\Rightarrow I_1=\underset{0}{\overset{\pi /2}{\int }}(\cos x+\cos 2x)dx$
$=1+0=1$