Question

The value of the definite integral $\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin 3x}{\sin x+\cos x}dx$ is equal to

• A. $1+\frac{\pi }{2}+\frac{\pi }{4}$
• B. $1-\frac{\pi }{4}-\frac{\pi }{2}$
• C. $1+\frac{\pi }{2}-\frac{\pi }{4}$
• D. $1-\frac{\pi }{2}+\frac{\pi }{4}$

Answer 1

Answer: (D)

$I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin 3x-\sin x}{\sin x+\cos x}dx+\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin x}{\sin x+\cos x}dx$
$I=I_1+I_2$
Now king and add in $I_2\Rightarrow 2I_2=\frac{\pi }{2}\Rightarrow I_2=\frac{\pi }{4}$
$I_1=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{2(\cos 2x)\sin x}{\sin x+\cos x}dx=2\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sin x(\cos x-\sin x)dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sin 2xdx-2\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{2}xdx$
${\therefore I_1=\frac{-1}{2}\cos 2x]}_0^{\frac{\pi }{2}}-2\cdot \frac{\pi }{4}=1-\frac{\pi }{2}$
$\therefore I=I_1+I_2=1-\frac{\pi }{2}+\frac{\pi }{4}$