Integrate a Function Using the Sine Case

When the function you’re integrating includes a term of the form (a2bx2)n, draw your trig substitution triangle for the sine case. For example, suppose that you want to evaluate the following integral:

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This is a sine case, because a constant minus a multiple of x2 is being raised to a power

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

  1. Draw the trig substitution triangle for the correct case.

    image2.jpg

    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:

    image3.png
  2. 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:

    image4.png
  3. 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 radical

    image5.png

    as the numerator and the constant 2 as the denominator. Then set this fraction equal to the appropriate trig function:

    image6.png

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

    image7.png

    this fraction is equal to

    image8.png

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

    image9.png

    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:

    image10.png

    This time, the numerator is the opposite side of the triangle and the denominator is the hypotenuse

    image11.png

    so this fraction is equal to

    image12.png

    Now solve for x and then differentiate:

    image13.png
  4. Rewrite the integral in terms of theta and evaluate it:

    image14.png
  5. To change those two theta terms into x terms, reuse the following equation:

    image15.png

    So here’s a substitution that gives you an answer:

    image16.png

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

image17.png

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

image18.png

Now use your trig substitution triangle to substitute values for

image19.png

in terms of x:

image20.png

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

image21.png
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