## University Calculus: Early Transcendentals (3rd Edition)

$$\int^{\pi/2}_0\sin^7ydy=\frac{16}{35}$$
$$A=\int^{\pi/2}_0\sin^7ydy$$ This is Case 1, so we need to find a way to rewrite $$(\sin^2y)^k\sin y=(1-\cos^2y)^k\sin y$$ We have, $$A=\int^{\pi/2}_0(\sin^2y)^3\sin ydy$$ $$A=-\int^{\pi/2}_0(1-\cos^2y)^3d(\cos y)$$ We would take $u=\cos y$ - For $y=0$, $u\cos0=1$ - For $y=\pi/2$, $u=\cos\pi/2=0$ $$A=-\int^0_1(1-u^2)^3du=\int^1_0(1-u^2)^3du$$ $$A=\int^1_0(1-3u^2+3u^4-u^6)du$$ $$A=\Big(u-u^3+\frac{3u^5}{5}-\frac{u^7}{7}\Big)\Big]^1_0$$ $$A=1-1^3+\frac{3\times1^5}{5}-\frac{1^7}{7}$$ $$A=\frac{3}{5}-\frac{1}{7}=\frac{16}{35}$$