How do you integrate #int sin^3x# by integration by parts method?

Question

Elijah Abram · Accepted Answer

#=1/3 cos^3 - cos x + C#

#int \ sin^3 x \ dx#

#= int \d/dx (-cos x) sin^2 x \ dx#

which by IBP

#=-cos x sin^2x + int \cos x d/dx( sin^2 x) \ dx#

#=-cos x sin^2 x + 2 int \cos^2 x sin x \ dx#

#=-cos x sin^2 x + 2 int \d/dx ( - 1/3 cos^3 x) \ dx#

#=-cos x sin^2 x - 2/3 cos^3 x + C#

#=-cos x (1 - cos^2 x) - 2/3 cos^3 x + C#

#=-cos x + cos^3 x - 2/3 cos^3 x + C#

#=1/3 cos^3 - cos x + C#

Paisley Johnson · Answer

undefined To integrate $ \int \sin^3(x) \, dx $ using integration by parts, you can use the following steps:

1. Start by rewriting $ \sin^3(x) $ as $ \sin^2(x) \cdot \sin(x) $.
2. Choose $ u = \sin^2(x) $ and $ dv = \sin(x) \, dx $.
3. Calculate $ du $ and $ v $ by taking derivatives and integrals, respectively.
4. Apply the integration by parts formula: $ \int u \, dv = uv - \int v \, du $.
5. Substitute the values of $ u $, $ du $, $ v $, and $ dv $ into the formula.
6. Evaluate the resulting integral.

Following these steps, you'll find that $ \int \sin^3(x) \, dx = -\frac{1}{3}\cos(x) + \frac{1}{3}\sin^3(x) + C $, where $ C $ is the constant of integration.