Question

Evaluate: ${\int }_0^{\frac{\pi }{4}}\frac{(sinx+cosx)}{9+16sin2x}dx$

Answer 1

Let $-\cos x+\sin x=t\Rightarrow (\sin x+\cos x)dx=dt$

${\cos }^{2}x+{\sin }^{2}x-2\sin x\cos x=t^2$

$\Rightarrow 1-\sin 2x=t^2\Rightarrow \sin 2x=1-t^2$

when $x=0,t=\cos 0+\sin 0=-1$

$x=\frac{\pi }{4},t=-\cos \frac{\pi }{4}+\sin \frac{\pi }{4}=-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=0$

$\therefore {\int }_0^{\pi /4}\frac{(\sin x+\cos x)}{(9+16\sin 2x)}dx={\int }_{-1}^0\frac{dt}{9+16(1-t{)}^2}$

$={\int }_{-1}^0\left(\frac{dt}{\frac{25}{16}-16t^2}\right)\frac{1}{16}=\frac{1}{16}{\int }_{-1}^0\frac{dt}{{\left(\frac{5}{4}\right)}^2-t^2}$

$=\frac{1}{16}{\int }_{-1}^0\left(\frac{dt}{(\frac{5}{4}-t)}(\frac{5}{4}+t)\right)-\frac{1}{16}{\int }_{-1}^0(\frac{1}{\frac{5}{4}+t}+\frac{1}{\frac{5}{4}-t})\frac{1}{2\left(\frac{5}{4}\right)}dt$

$=\frac{1}{16}.\frac{2}{5}{\int }_{-1}^0\frac{1}{\frac{5}{4}+t}+\frac{1}{\frac{5}{4}-t}dt$

$=\frac{1}{40}{[\ln (\frac{5}{4}+t)-\ln (\frac{5}{4}-t)]}_{-1}^0$

$=\frac{1}{40}\ln \left(\frac{\frac{5}{4}+t}{\frac{5}{4}-t}\right){\int }_{-1}^0=\frac{1}{40}[\ln \left(\frac{5/4}{5/4}\right)-\ln \left(\frac{(\frac{5}{4}-1)}{\frac{5}{4}+1}\right)]$

$=\frac{1}{40}[\ln -1-\ln \left(\frac{1/4}{9/4}\right)]=\frac{-1}{40}\ln \left(\frac{1}{9}\right)$

$=\frac{-1}{40}\left[\ln \right(3{)}^{-2}]=\frac{-2}{-40}{\ln }^3=\frac{1}{20}{\ln }^3$