# Trigonometric Integral

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.

## Answers and Replies

Homework Helper
Gold Member
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.

Mentor
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.