Question

Evaluate: ${\int }_0^{\frac{\pi }{4}}\left(\frac{sinx+cosx}{3+sin2x}\right)dx.$

Answer 1

Let I = ${\int }_0^{\frac{\pi }{4}}\left(\frac{sinx+cosx}{3+sin2x}\right)dx.$ Put sin x - cos x = t $\Rightarrow$ (sin x + cos x) dx = dt

And, $(sinx-cosx{)}^2=t^2\Rightarrow I-t^2=sin2x.$ Also when x = 0 $\Rightarrow$ t = -1 and when x = $\frac{\pi }{4}\Rightarrow t=0$

$\therefore I={\int }_0^{-1}\frac{dt}{4-t^2}\Rightarrow I=\frac{1}{2\left(2\right)}{\left[log\left|\begin{array}{c}\frac{2+t}{2-t}\end{array}\right|\right]}_{-1}^0\Rightarrow I=\frac{1}{4}[log\left|\begin{array}{c}\frac{2+0}{2-0}\end{array}\right|-log\left|\begin{array}{c}\frac{2-1}{2+1}\end{array}\right|]$

$\Rightarrow I=\frac{1}{4}(0+log3)\therefore I=\frac{log3}{4}$.