Question

The value of $\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^{3}x}{\sin x+\cos x}dx$ is:

• A. $\frac{\pi -1}{2}$
• B. $\frac{\pi -1}{4}$
• C. $\frac{\pi -2}{8}$
• D. $\frac{\pi -2}{4}$

Answer 1

The correct option is B $\frac{\pi -1}{4}$

$I=\underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^{3}x}{\sin x+\cos x}dx\cdots \left(i\right)Also,I=\underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^3(\frac{\pi }{2}-x)}{\sin (\frac{\pi }{2}-x)+\cos (\frac{\pi }{2}-x)}dx=\underset{0}{\overset{\pi /2}{\int }}\frac{{\cos }^{3}x}{\sin x+\cos x}dx\cdots \left(ii\right)$
On adding $\left(i\right)and\left(ii\right)$ we get
$2I=\underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^{3}x+{\cos }^{3}x}{\sin x+\cos x}dx\Rightarrow I=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^{3}x+{\cos }^{3}x}{\sin x+\cos x}dx\Rightarrow I=\frac{1}{2}\underset{0}{\overset{\pi /2}{\int }}(1-\sin x\cos x)dx=\frac{1}{2}[x-\frac{{\sin }^{2}x}{2}{]}_0^{\pi /2}$
$=\frac{\pi }{4}-\frac{1}{4}=\frac{\pi -1}{4}$