a m p ; I = 80 ∫ 0 π 4 ( sin θ + cos θ 9 + 16 ( 2 sin θ ⋅ cos θ ) ) d θ a m p ; = 80 ∫ 0 π 4 sin θ + cos θ 9 − 16 ( 1 − 2 sin θ ⋅ cos θ − 1 ) d θ \begin{aligned}& 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(1-2 \sin \theta \cdot \cos \theta-1)} d \theta\end{aligned} am p ; I = 80 ∫ 0 4 π ( 9 + 16 ( 2 sin θ ⋅ cos θ ) sin θ + cos θ ) d θ am p ; = 80 ∫ 0 4 π 9 − 16 ( 1 − 2 sin θ ⋅ cos θ − 1 ) sin θ + cos θ d θ
a m p ; = 80 ∫ 0 π 4 sin θ + cos θ 9 + 16 − 16 ( sin θ − cos θ ) 2 d θ a m p ; Let sin θ − cos θ = t a m p ; ( cos θ + sin θ ) d θ = d t a m p ; = 80 ∫ − 1 0 d t 25 − 16 t 2 a m p ; = 80 16 ∫ − 1 0 d t ( 5 4 ) 2 − t 2 a m p ; = 5 2 ( 5 4 ) ln ∣ 5 4 + t 5 4 − t ∣ ] − 1 0 a m p ; = 2 ln ( 1 ) + 4 ln 3 a m p ; = 4 ln 3 \begin{aligned}&=80 \int_0^{\frac{\pi}{4}} \frac{\sin \theta+\cos \theta}{9+16-16(\sin \theta-\cos \theta)^2} \mathrm{~d} \theta\\&\text { Let } \sin \theta-\cos \theta=\mathrm{t}\\&(\cos \theta+\sin \theta) \mathrm{d} \theta=\mathrm{dt}\\&=80 \int_{-1}^0 \frac{\mathrm{dt}}{25-16 \mathrm{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\\&=2 \ln (1)+4 \ln 3\\&=4 \ln 3\end{aligned} am p ; = 80 ∫ 0 4 π 9 + 16 − 16 ( sin θ − cos θ ) 2 sin θ + cos θ d θ am p ; Let sin θ − cos θ = t am p ; ( cos θ + sin θ ) d θ = dt am p ; = 80 ∫ − 1 0 25 − 16 t 2 dt am p ; = 16 80 ∫ − 1 0 ( 4 5 ) 2 − t 2 dt am p ; = 2 ( 4 5 ) 5 ln 4 5 − t 4 5 + t ] − 1 0 am p ; = 2 ln ( 1 ) + 4 ln 3 am p ; = 4 ln 3