Answer to Question #131176 in Calculus for miya

Question

1. ∫sin^43xcos^23xdx
2. ∫x^5 sqrt of 1-x^3 dx

Answer 1

$∫sin^4(3x)cos^2(3x)dx=∫sin^2(3x)* (sin^2(3x)*cos^2(3x))dx=∫(1/2)*(1-cos(6x))*(1/4)*sin^2(6x) dx=∫(1/8)*(1-cos(6x))*(1/2)*(1-cos(12x))dx=∫(1/16)*(1-cos(6x)-cos(12x)+ cos(6x)*cos(12x))dx=∫(1/16)*(1-cos(6x)-cos(12x)+ cos(6x)*cos(12x))dx=∫(1/16)*(1-cos(6x)-cos(12x)+ (1/2)*cos(6x)+(1/2)*cos(18x))dx=∫(1/16)*(1-(1/2)*cos(6x)-cos(12x)+(1/2)*cos(18x))dx=(1/16)*(x-(1/12)*sin(6x)-(1/12)*sin(12x)+(1/36)*sin(18x))+C=x/16-(1/192)*sin(6x)-(1/192)*sin(12x)+(1/576)*sin(18x)+C$$∫x^5* (1-x^3)^(1/2) dx=∫x^3*x^2*(1-x^3)^(1/2) dx=∫(1-u)*u^(1/2) d(u/(-3))=-1/3∫u^(1/2)-u^(3/2) du=-1/3(u^(3/2)/(3/2)-u^(5/2)/(5/2))+C=-(2/9)*u^(3/2)+(2/15)*u^(5/2)+C=-(2/9)* (1-x^3)^(3/2)+(2/15)*(1-x^3)^(5/2)+C$