Answer to Question #139776 in Calculus for Moel Tariburu

Question

Evaluate the integral ∫▒(〖cos〗^5 (lnx) 〖sin〗^2 (lnx))/x dx

Answer 1

$\int \frac{cos^5(lnx)sin^2(lnx)}{x}dx=\int cos^5(lnx)sin^2(lnx)d(lnx)=\int cos(lnx)cos^4(lnx)sin^2(lnx)d(lnx)=$

$=\int cos^4(lnx)sin^2(lnx)d(sin(lnx))=|t=sin(lnx), cos^2(lnx)=1-sin^2(lnx)|=$

$=\int (1-t^2)^2t^2dt=\int(1-2t^2+t^4)t^2dt=\int(t^6-2t^4+t^2)dt=\frac17t^7-\frac25t^5+\frac13t^3+C=$ $=\frac17sin^7(lnx)-\frac25sin^5(lnx)+\frac13sin^3(lnx)+C$