Solving Trigonometric Integrals: \int sin^6(x)cos^3(x) dx

[tangibleLime]

Homework Statement

$$\int sin^6(x)cos^3(x) dx$$

Homework Equations

$$cos^2(x) = 1-sin^2(x)$$

The Attempt at a Solution

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

$$\int sin^6(x)cos^3(x) dx$$
$$\int sin^6(x)(1-sin^2(x))cos(x) dx$$

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

$$\int u^6(1-u^2) du$$
$$\int u^6-u^8) du$$
$$\frac{1}{7}u^7-\frac{1}{9}u^9$$

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

$$\frac{1}{7}sin(x)^7-\frac{1}{9}sin(x)^9 + C$$

This was found to be incorrect.

 

[fzero]

You have the correct answer. There are various trig identities that could be applied to this, so it's possible the answer you're trying to compare to is equivalent.

 

[Mark44]

You can verify that your answer is correct by differentiating it, which should get you back to your integrand. For this problem, the derivative of your answer is sin⁶(x)cos(x) - sin⁸(x)cos(x) = sin⁶(x)cos(x) (1 - sin²(x)) = sin⁶(x)cos³(x), which is the same as your integrand.