Mathematics - Definite Integration Question with Solution | TestHub

$\begin{array}{rl}& I={\int }_0^{\frac{\pi }{4}}\left(\frac{\sin \theta +\cos \theta }{9-16\sin 2\theta }\right)d\theta \\ & \text{ Take }\sin \theta -\cos \theta =t\\ & (\cos \theta +\sin \theta )d\theta =dt\\ & (\sin \theta -\cos \theta {)}^2=t^2\\ & \Rightarrow \sin 2\theta =1-t^2\\ & \theta =0\to t=-1\\ & \theta =\frac{\pi }{4}\to t=0\\ & I={\int }_{-1}^0\frac{dt}{9+16\left(1-t^2\right)}\\ & =\frac{1}{16}{\int }_{-1}^0\frac{dt}{25}\frac{16}{16}-t^2\\ & =\frac{1}{4}{\left[\frac{1}{10}{\log }_{|\frac{5+4t}{}}^{5-4t}\right]}_{-1}^0\\ & =\frac{1}{40}\left[0+{\log }_{e}9\right]\\ & I=\frac{{\log }_{e}9}{40}\\ & 80I=2{\log }_{e}9\\ & 80I=4{\log }_{e}3\end{array}$