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## Homework Statement

[tex]\int sin^6(x)cos^3(x) dx[/tex]

## Homework Equations

[tex]cos^2(x) = 1-sin^2(x)[/tex]

## The Attempt at a Solution

Since [tex]cos[/tex] has an odd power, I took one out to make it [tex]cos^2(x)[/tex], which can be used in the identity above.

[tex]\int sin^6(x)cos^3(x) dx[/tex]

[tex]\int sin^6(x)(1-sin^2(x))cos(x) dx[/tex]

I substituted [tex]u = sin(x)[/tex] since [tex]du = cos(x)[/tex] will take care of the right side of that integral.

[tex]\int u^6(1-u^2) du[/tex]

[tex]\int u^6-u^8) du[/tex]

[tex]\frac{1}{7}u^7-\frac{1}{9}u^9[/tex]

Then I put sin(x) back, replacing the u's and added the constant of integration.

[tex]\frac{1}{7}sin(x)^7-\frac{1}{9}sin(x)^9 + C[/tex]

This was found to be incorrect.