The integral 80 ∫_0^ π 4 ( sin θ+cos θ 9+16 sin 2 θ ) d θ is equal to :

$$\begin{aligned} & I=\int_0^{\frac{\pi}{4}}\left(\frac{\sin \theta+\cos \theta}{9-16 \sin 2 \theta}\right) d \theta \\ & \text { Take } \sin \theta-\cos \theta=t \\ & (\cos \theta+\sin \theta) d \theta=d t \\ & (\sin \theta-\cos \theta)^2=t^2 \\ & \Rightarrow \sin 2 \theta=1-t^2 \\ & \theta=0 \rightarrow t=-1 \\ & \quad \theta=\frac{\pi}{4} \rightarrow t=0 \\ & I=\int_{-1}^0 \frac{d t}{9+16\left(1-t^2\right)} \\ & =\frac{1}{16} \int_{-1}^0 \frac{d t}{25} \frac{16}{16}-t^2 \\ & =\frac{1}{4}\left[\frac{1}{10} \log _{\left\lvert\, \frac{5+4 t}{}\right.}^{5-4 t}\right]_{-1}^0 \\ & =\frac{1}{40}\left[0+\log _e 9\right] \\ & I=\frac{\log _e 9}{40} \\ & 80 I=2 \log _e 9 \\ & 80 I=4 \log _e 3\end{aligned}$$​I=∫04π​​(9−16sin2θsinθ+cosθ​)dθ Take sinθ−cosθ=t(cosθ+sinθ)dθ=dt(sinθ−cosθ)2=t2⇒sin2θ=1−t2θ=0→t=−1θ=4π​→t=0I=∫−10​9+16(1−t2)dt​=161​∫−10​25dt​1616​−t2=41​[101​log∣5+4t​5−4t​]−10​=401​[0+loge​9]I=40loge​9​80I=2loge​980I=4loge​3​