Question

Solve:${\int }_0^{\pi /2}\frac{\sin xdx}{(\sin x+\cos x{)}^3}$

Answer 1

Solution:

${\int }_0^{\frac{\pi }{2}}\frac{sinxdx}{(sinx+cosx{)}^3}=I$

using properties of definite integrals

$I={\int }_0^{\frac{\pi }{2}}\frac{sin(\frac{\pi }{2}+0-\pi )}{\left(sin\right(\frac{\pi }{2}+0-x)+cos(\frac{\pi }{2}+0-x){)}^3}$

$={\int }_0^{\frac{\pi }{2}}\frac{cosx}{(sinx+cosx{)}^3}$

$2I={\int }_0^{\frac{\pi }{2}}\frac{(sinx+cosx)}{(sinx+cosx{)}^3}$

$={\int }_0^{\frac{\pi }{2}}\frac{1}{(sinx+cosx{)}^2}$

$={\int }_0^{\frac{\pi }{2}}\frac{1}{1+sin2x}$

$={\int }_0^{\frac{\pi }{2}}\frac{1}{1+cos(\frac{\pi }{2}-2x)}$

$={\int }_0^{\frac{\pi }{2}}1/2co{s}^2(\frac{\pi }{4}-\frac{2x}{2})$

$={\int }_0^{\frac{\pi }{2}}\frac{1}{2}se{c}^2(\frac{\pi }{4}-\frac{2x}{2})=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}tan(\frac{\pi }{4}-x)$

$=-1$