The integral \(80{\int }_{0}^{\frac{\pi }{4}}\left(\frac{\sin \theta +\cos \theta }{9+16\sin 2\theta }\right)d\theta\) i…

\(\text{Let}I=80{\int }_{0}^{\frac{\pi }{4}}\left(\frac{\sin \theta +\cos \theta }{9+16(2\sin \theta \cdot \cos \theta )}\right)d\theta \\ =80{\int }_{0}^{\frac{\pi }{4}}\frac{\sin \theta +\cos \theta }{9+16-16(\sin \theta -\cos \theta {)}^{2}}d\theta \\ \text{Put}\sin \theta -\cos \theta =t\\ \left(\cos \theta +\sin \theta \right)\mathrm{d}\theta =dt\\ I=80{\int }_{-1}^{0}\frac{\mathrm{dt}}{25-16{t}^{2}}\\ =\frac{80}{16}{\int }_{-1}^{0}\frac{\mathrm{dt}}{{\left(\frac{5}{4}\right)}^{2}-{\mathrm{t}}^{2}}{\left.=\frac{5}{2\left(\frac{5}{4}\right)}\ln \left(\frac{\frac{5}{4}+t}{\frac{5}{4}-t}\right)\right]}_{-1}^{0}\\ =4\ln 3\)