Following the hint, we wish to perform the substitution u = sin ( x ) {\displaystyle u=\sin(x)} , but we must first rewrite the integrand by using the fundamental identity sin 2 ( x ) + cos 2 ( x ) = 1 {\displaystyle \sin ^{2}(x)+\cos ^{2}(x)=1} :
∫ 0 π / 2 sin 4 ( x ) cos 3 ( x ) d x = ∫ 0 π / 2 sin 4 ( x ) ( 1 − sin 2 ( x ) ) cos ( x ) d x = ∫ 0 π / 2 ( sin 4 ( x ) − sin 6 ( x ) ) cos ( x ) d x . {\displaystyle \int _{0}^{\pi /2}\sin ^{4}(x)\cos ^{3}(x)\mathrm {d} x=\int _{0}^{\pi /2}\sin ^{4}(x)\left(1-\sin ^{2}(x)\right)\cos(x)\mathrm {d} x=\int _{0}^{\pi /2}\left(\sin ^{4}(x)-\sin ^{6}(x)\right)\cos(x)\mathrm {d} x.}
We now perform the u {\displaystyle u} -substitution above. Remember to change the integration bounds for the integral in terms of u {\displaystyle u} :
∫ 0 π / 2 sin 4 ( x ) cos 3 ( x ) d x = ∫ sin ( 0 ) sin ( π / 2 ) ( u 4 − u 6 ) d u = [ u 5 5 − u 7 7 ] 0 1 = 1 5 − 1 7 − 0 + 0 = 2 35 . {\displaystyle \int _{0}^{\pi /2}\sin ^{4}(x)\cos ^{3}(x)\mathrm {d} x=\int _{\sin(0)}^{\sin(\pi /2)}(u^{4}-u^{6})\mathrm {d} u=\left[{\frac {u^{5}}{5}}-{\frac {u^{7}}{7}}\right]_{0}^{1}={\frac {1}{5}}-{\frac {1}{7}}-0+0={\frac {2}{35}}.}
, but we must first rewrite the integrand by using the fundamental identity 
:
-substitution above. Remember to change the integration bounds for the integral in terms of :
![{\displaystyle \int _{0}^{\pi /2}\sin ^{4}(x)\cos ^{3}(x)\mathrm {d} x=\int _{\sin(0)}^{\sin(\pi /2)}(u^{4}-u^{6})\mathrm {d} u=\left[{\frac {u^{5}}{5}}-{\frac {u^{7}}{7}}\right]_{0}^{1}={\frac {1}{5}}-{\frac {1}{7}}-0+0={\frac {2}{35}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/e07ce94cd54aac56442133b52aa8a10120df724c)
