How do you integrate #sin^5 (x) * cos^3 (x)#?

1 Answer
Jun 24, 2016

Use one of the substitutions: #u=sinx# #" "# OR #" "# #u=cosx#

Explanation:

#int sin^5x cos^3x dx = int sin^5xcos^2xcosx dx#

# = int sin^5x(1-sin^2x)cosx dx#

# = int sin^5x cosx dx - int sin^7xcosx dx#

Substitute #u = sinx# to get

# = 1/6 sin^6x - 1/8 sin^8x +C#

OR

#int sin^5x cos^3x dx = int sin^4xcos^3xsinx dx#

# = int (sin^2x)^2 cos^3x sinx dx#

# = int (1-cos^2x)^2 cos^3x sinx dx#

# = int (1-2cos^2x+cos^4x) cos^3x sinx dx#

# = int cos^3x sinx dx - int 2cos^5x sinx dx + int cos^7x sinx dx#

Substitute #u = cosx#, so #du = -sinx# to get

# = -1/4cos^4x+1/3cos^6x-1/8cos^8x +C#