Question

$\int_{0}^{\pi} \sin^{5}x \cos^{4}x \,dx$ equals to

• A. $\frac{16}{315}$
• B. $\frac{8}{315}$
• C. $\frac{8\pi}{315}$
• D. $\frac{16\pi}{315}$

Answer 1

Answer: (A)

$\frac{16}{315}$

Answer 2

Let $I = \int_{0}^{\pi} \sin^{5}x \cos^{4}x \,dx$. We use the property $\int_{0}^{a} f(x) \,dx = \int_{0}^{a} f(a-x) \,dx$. Let $f(x) = \sin^{5}x \cos^{4}x$. Then $f(\pi-x) = \sin^{5}(\pi-x) \cos^{4}(\pi-x) = (\sin x)^{5} (-\cos x)^{4} = \sin^{5}x \cos^{4}x = f(x)$. Thus, $I = \int_{0}^{\pi} \sin^{5}x \cos^{4}x \,dx = 2 \int_{0}^{\pi/2} \sin^{5}x \cos^{4}x \,dx$. Let $u = \cos x$, then $du = -\sin x \,dx$. When $x=0$, $u=1$. When $x=\pi/2$, $u=0$. $\sin^{5}x \cos^{4}x = \sin^{4}x \cos^{4}x \sin x = (1-\cos^{2}x)^{2} \cos^{4}x \sin x = (1-u^{2})^{2} u^{4} (-du)$. So, $I = 2 \int_{1}^{0} (1-u^{2})^{2} u^{4} (-du) = 2 \int_{0}^{1} (1-2u^{2}+u^{4}) u^{4} \,du$ $I = 2 \int_{0}^{1} (u^{4} - 2u^{6} + u^{8}) \,du = 2 \left[ \frac{u^{5}}{5} - \frac{2u^{7}}{7} + \frac{u^{9}}{9} \right]_{0}^{1}$ $I = 2 \left( \frac{1}{5} - \frac{2}{7} + \frac{1}{9} \right) = 2 \left( \frac{63 - 90 + 35}{315} \right) = 2 \left( \frac{8}{315} \right) = \frac{16}{315}$.