integration of cos5x•sin2x
Answers
To evaluate
are positive integers and at least one of
Pull off one from the odd power. (If both are odd, it is simpler, but not necessary, to make the lesser power even.)
Change the remaining even power to the other function using
How do you integrate #sin^2(x) cos^5(x)#? CalculusIntegration by Parts 1 Answer Jim H Oct 29, 2015 Recommended: learn this pattern. Explanation: To evaluate #int sin^mxcos^nx dx# where #m,n# are positive integers and at least one of #m,n# is odd. Pull off one from the odd power. (If both are odd, it is simpler, but not necessary, to make the lesser power even.) Change the remaining even power to the other function using #sin^2x+cos^2x = 1#. Expand and substitute to get a polynomial in #u# It may sound complicated, but here's how it works for this question: #intsin^2xcos^5xdx=intsin^2xcos^4xunderbrace(cosxdx)_(du)# We are going to use #u = sinx# so we need to rewrite #cos^4x = (cos^2x)^2 = (1-sin^2x)^2 = 1-2sinx+sin^2x# Our integral becomes: #intsin^2xunderbrace(( 1-2sinx+sin^2x))_(cos^4x) cosx dx# Answer link Related topic Integration by Parts Questions Related questions How do I find the integral #int(x*ln(x))dx# ? How do I find the integral #int(cos(x)/e^x)dx# ? How do I find the integral #int(x*cos(5x))dx# ? How do I find the integral #int(x*e^-x)dx# ? How do I find the integral #int(x^2*sin(pix))dx# ? How do I find the integral #intln(2x+1)dx# ? How do I find the integral #intsin^-1(x)dx# ? How do I find the integral #intarctan(4x)dx# ? How do I find the integral #intx^5*ln(x)dx# ? How do I find the integral #intx*2^xdx# ?