How do I evaluate the indefinite integral
#intsin^3(x)*cos^2(x)dx# ?

Question

How do I evaluate the indefinite integral
 #intsin^3(x)*cos^2(x)dx# ?

Luke Alvarez · Accepted Answer

The answer is #-(cos^3x)/3+(cos^5x)/5+C#. The trick with sinusoidal powers is to use identities so that you can have #sin x# or #cos x# with a power of 1 and use substitution. In this case, it is easier to get #sin x# to a power of 1 using #sin^2x=1-cos^2x#. #int sin^3x*cos^2x dx#
#=int sin x(1-cos^2x)cos^2x dx#
#=int sin x(cos^2x-cos^4x)dx#
#=int sin x cos^2xdx-int sin x cos^4x dx# It's time to use substitution now: #u=cos x#
#du = -sin x dx# #int sin x cos^2xdx-int sin x cos^4x dx#
#=int -u^2 du+ int u^4 du#
#=-(u^3)/3+(u^5)/5+C#
#=-(cos^3x)/3+(cos^5x)/5+C#

Emily Ammerman · Answer

undefined To evaluate the indefinite integral ∫sin^3(x)*cos^2(x)dx, you can use trigonometric identities to simplify the integral.

Start by using the identity sin^2(x) + cos^2(x) = 1 to rewrite cos^2(x) as 1 - sin^2(x). Then, substitute this expression into the integral.

You'll end up with an integral involving sin^3(x) and sin(x), which can be simplified using integration by parts. Let u = sin^2(x) and dv = sin(x)dx, then apply integration by parts to find the integral.

After integrating by parts, you should obtain the solution to the indefinite integral.

How do I evaluate the indefinite integral #intsin^3(x)*cos^2(x)dx# ?