Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises - Page 498: 3

$\frac{1}{{210}}$

$$\eqalign{ & \text{Let }I=\int_0^{\pi /2} {{{\cos }^9}x{{\sin }^5}x} dx \cr & {\text{Split }}{\sin ^5}x{\text{ as si}}{{\text{n}}^4}x\sin x \cr & I= \int_0^{\pi /2} {{{\cos }^9}x{\text{si}}{{\text{n}}^4}x\sin x} dx \cr & {\text{Write si}}{{\text{n}}^4}x{\text{ as }}{\left( {{{\sin }^2}x} \right)^2} \cr & I= \int_0^{\pi /2} {{{\cos }^9}x{{\left( {{{\sin }^2}x} \right)}^2}\sin x} dx \cr & {\text{using }}{\sin ^2}x + {\cos ^2}x = 1 \cr & I= \int_0^{\pi /2} {{{\cos }^9}x{{\left( {1 - {{\cos }^2}x} \right)}^2}\sin x} dx \cr & {\text{Expand the binomial}} \cr & I= \int_0^{\pi /2} {{{\cos }^9}x\left( {1 - 2{{\cos }^2}x + {{\cos }^4}x} \right)\sin x} dx \cr & \cr & {\text{Integrate by the substitution method}} \cr & {\text{Let }}u = \cos x,{\text{ }}du = - \sin xdx,{\text{ }} - du = \sin xdx \cr & {\text{The new limits of integration are:}} \cr & x = \frac{\pi }{2} \to u = \cos \left( {\frac{\pi }{2}} \right) = 0 \cr & x = 0 \to u = \cos \left( 0 \right) = 1 \cr & {\text{Therefore}} \cr & I= \int_1^0 {{u^9}\left( {1 - 2{u^2} + {u^4}} \right)\left( { - 1} \right)} du \cr & = \int_0^1 {{u^9}\left( {1 - 2{u^2} + {u^4}} \right)} du \cr & = \int_0^1 {\left( {{u^9} - 2{u^{11}} + {u^{13}}} \right)} du \cr & {\text{Use the power rule }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr &I = \left[ {\frac{{{u^{10}}}}{{10}} - \frac{{{u^{12}}}}{6} + \frac{{{u^{14}}}}{{14}}} \right]_0^1 \cr & = \left[ {\frac{{{{\left( 1 \right)}^{10}}}}{{10}} - \frac{{{{\left( 1 \right)}^{12}}}}{6} + \frac{{{{\left( 1 \right)}^{14}}}}{{14}}} \right] - \left[ {\frac{{{{\left( 0 \right)}^{10}}}}{{10}} - \frac{{{{\left( 0 \right)}^{12}}}}{6} + \frac{{{{\left( 0 \right)}^{14}}}}{{14}}} \right] \cr & {\text{Simplifying}} \cr & I = \frac{1}{{210}} \cr} $$