Question

Evaluate the definite integrals.
${\int }_0^{\frac{\pi }{4}}\frac{sinxcosx}{co{s}^{4}x+si{n}^{4}x}dx$

Answer 1

Let $I={\int }_0^{\frac{\pi }{4}}\frac{sinxcosx}{co{s}^{4}x+si{n}^{4}x}dx={\int }_0^{\frac{\pi }{4}}\frac{\sin xcosx}{(1-si{n}^{2}x)+(si{n}^{2}x{)}^2}[\because co{s}^{2}x=1-si{n}^{2}x]$
Put $si{n}^{2}x=t\to 2sinxcosxdx=dt$
for limit when $x=0\Rightarrow t=0andwhenx=\frac{\pi }{4}\Rightarrow t=\frac{1}{2}\therefore I={\int }_0^{\frac{1}{2}}\frac{sinxcosx}{(1-t{)}^2+t^{2}2sinxcosx}dt\frac{1}{2}{\int }_0^{\frac{1}{2}}\frac{1}{(1-t{)}^2+t^2}dt=\frac{1}{2}{\int }_0^{\frac{1}{2}}\frac{1}{2t^2-2t+1}dt=\frac{1}{2}.\frac{1}{2}{\int }_0^{\frac{1}{2}}\frac{1}{t^2-t+\frac{1}{2}}dt=\frac{1}{4}{\int }_0^{\frac{1}{2}}\frac{1}{t^2-t+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2+\frac{1}{2}}dt=\frac{1}{4}{\int }_0^{\frac{1}{2}}\frac{1}{{(t-\frac{1}{2})}^2+{\left(\frac{1}{2}\right)}^2}dt=\frac{1}{4}.\frac{1}{\frac{1}{2}}{\left[ta{n}^{-1}\left(\frac{t-\frac{1}{2}}{\frac{1}{2}}\right)\right]}_0^{\frac{1}{2}}=\frac{1}{2}[ta{n}^{-1}0-ta{n}^{-1}(-1\left)\right]=\frac{1}{2}.\frac{\pi }{4}=\frac{\pi }{8}$