# How do you evaluate #\int ( \sin x + \cos x ) ^ { 2} d x #?

##### 1 Answer

or

#### Explanation:

You can't tackle this integral as it is. To solve it, you'll need to break it down into something you can work with. The best course of action is to foil out the

Now, there's something we can work with. Recall that

Applying our identity:

- Set
#u = "some value"# . - Set it to something that will cancel out other
#x# terms in the integral. - Take the derivative of both sides of that equation with respect to
#x# . - Solve for
#dx# in terms of something#"* du# - Plug that in for
#dx# in your original integral.

Note that you can pick *either*

Let

Plug back in:

Now, we just evaluate the indefinite integral in terms of

Since we started with

Now we tag on the answer from the other part of the integral:

And that's your final answer. If you want to check if it is correct, just take a derivative, and you should be able to get back to your original expression.

Also note that if you had substituted

Hope that helped :)