The integral $80 \int_0^{\frac{\pi}{4}}\left(\frac{\sin \theta+\cos \theta}{9+16 \sin 2 \theta}\right) d \theta$ is equal to :

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Aicharya · Accepted Answer

$\begin{aligned} & I=\int_0^{\frac{\pi}{4}}\left(\frac{\sin 	heta+\cos 	heta}{9-16 \sin 2 	heta}ight) d 	heta \ & 	ext { Take } \sin 	heta-\cos 	heta=t \ & (\cos 	heta+\sin 	heta) d 	heta=d t \ & (\sin 	heta-\cos 	heta)^2=t^2 \ & \Rightarrow \sin 2 	heta=1-t^2 \ & 	heta=0 ightarrow t=-1 \ & \quad 	heta=\frac{\pi}{4} ightarrow t=0 \ & I=\int_{-1}^0 \frac{d t}{9+16\left(1-t^2ight)} \ & =\frac{1}{16} \int_{-1}^0 \frac{d t}{25} \frac{16}{16}-t^2 \ & =\frac{1}{4}\left[\frac{1}{10} \log _{\left\lvert\, \frac{5+4 t}{}ight.}^{5-4 t}ight]_{-1}^0 \ & =\frac{1}{40}\left[0+\log _e 9ight] \ & I=\frac{\log _e 9}{40} \ & 80 I=2 \log _e 9 \ & 80 I=4 \log _e 3\end{aligned}$

The integral \(80 \int_0^{\frac{\pi}{4}}\left(\frac{\sin \theta+\cos \theta}{9+16 \sin 2 \theta}\right) d…

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