# Integrate a Function Using the Sine Case

When the function you’re integrating includes a term of the form (*a*^{2} – *bx*^{2})* ^{n}*, draw your trig substitution triangle for the

*sine case.*For example, suppose that you want to evaluate the following integral:

This is a sine case, because a constant minus a multiple of *x*^{2} is being raised to a power

Here’s how you use trig substitution to handle the job:

Draw the trig substitution triangle for the correct case.

This figure shows you how to fill in the triangle for the sine case. Notice that the radical goes on the

*adjacent*side of the triangle. Then, to fill in the other two sides of the triangle, you use the square roots of the two terms inside the radical — that is, 2 and*x**.*Place 2 on the hypotenuse and*x*on the opposite side.You can check to make sure that this placement is correct by using the Pythagorean theorem:

Identify the separate pieces of the integral (including

*dx*) that you need to express in terms of theta*.*In this case, the function contains two separate pieces that contain

*x**:*Express these pieces in terms of trig functions of theta

*.*This is the real work of trig substitution, but when your triangle is set up properly, this work becomes a lot easier. In the sine case,

*all*trig functions should be sines and cosines.To represent the radical portion as a trig function of theta

*,*first build a fraction using the radicalas the numerator and the constant 2 as the denominator. Then set this fraction equal to the appropriate trig function:

Because the numerator is the adjacent side of the triangle and the denominator is the hypotenuse

this fraction is equal to

Now a little algebra gets the radical alone on one side of the equation:

Next, you want to express

*dx*as a trig function of theta*.*To do so, build another fraction with the variable*x*in the numerator and the constant 2 in the denominator. Then set this fraction equal to the correct trig function:This time, the numerator is the opposite side of the triangle and the denominator is the hypotenuse

so this fraction is equal to

Now solve for

*x*and then differentiate:Rewrite the integral in terms of theta and evaluate it:

To change those two theta terms into

*x*terms, reuse the following equation:So here’s a substitution that gives you an answer:

This answer is perfectly valid so, technically speaking, you can stop here. However, some professors frown upon the nesting of trig and inverse trig functions, so they’ll prefer a simplified version of

To find this, start by applying the double-angle sine formula to

Now use your trig substitution triangle to substitute values for

in terms of *x:*

To finish, substitute this expression for that problematic second term to get your final answer in a simplified form: