## Calculus: Early Transcendentals 8th Edition

$$\int^{\pi/2}_0\sin^7\theta\cos^5\theta d\theta=\frac{1}{120}$$
$$A=\int^{\pi/2}_0\sin^7\theta\cos^5\theta d\theta$$ Here both the power of $\sin$ and $\cos$ are odd, so we can change any of each in terms of the other. Here we would save 1 $\sin$ factor and change the rest in terms of $\cos$ according to the formula $\sin^2\theta=1-\cos^2\theta$ Therefore, we have $$A=\int^{\pi/2}_0\sin^6\theta\cos^5\theta\sin \theta d\theta$$ $$A=\int^{\pi/2}_0(\sin^2\theta)^3\cos^5\theta\sin \theta d\theta$$ $$A=\int^{\pi/2}_0\cos^5\theta(1-\cos^2\theta)^3\sin\theta d\theta$$ Now we substitute $u=\cos\theta$ That makes $du=-\sin\theta d\theta$, or $\sin\theta d\theta=-du$ Also, for $\theta=0$, $u=1$ and for $\theta=\frac{\pi}{2}$, $u=0$ So, $$A=-\int^{0}_1 u^5(1-u^2)^3du$$ $$A=-\int^{0}_1 u^5(1-3u^2+3u^4-u^6)du$$ $$A=-\int^{0}_1 (u^5-3u^7+3u^9-u^{11})du$$ $$A=\int^0_1 (u^{11}-3u^9+3u^7-u^5)du$$ $$A=[\frac{u^{12}}{12}-\frac{3u^{10}}{10}+\frac{3u^8}{8}-\frac{u^6}{6}]^{0}_1$$ $$A=0-[\frac{1}{12}-\frac{3\times1}{10}+\frac{3\times1}{8}-\frac{1}{6}]$$ $$A=\frac{1}{120}$$