Trigonometric Substitutions

 

     We have already dealt with substitutions. Substitutions allow us to solve complicated-looking integrals by converting them into something more manageable.

 

     Now, we shall examine a different type of substitution: a trigonometric substitution, which will allow us to integrate more functions.

 

     Before we begin, though, let us recall some useful trigonometric identities.

 

     The first is the Pythagorean Identity: sin²θ + cos²θ = 1. If we divide all terms by cos²θ, then we have: . This is another identity which we will use later on.

 

Example 1: Evaluate .

 

     Notice that we cannot use a u-substitution.

 

          If we let u = 1  x², then du = -2xdx, but we don’t have -2xdx.

 

     Also, integration by parts will not work.

 

          If we let u = (1  x²)^-1/2, then du = x(1  x²)^-3/2dx. dv = dx, and so v = x.

 

          Putting it together, we have: , which has only made the integral worse.

 

     But what happens if we let x = sinθ. Notice that the quantity inside of the square root becomes 1  sin²θ = cos²θ, by the Pythagorean Identity above.

 

     Also, notice that dx = cosθdθ. Unlike before, here we do not have to solve for dx; it is already given to use.

 

     So, now we just plug everything in and solve.

 

      

 

     The question remains, what is θ? We need to solve for θ in terms of x.

 

     If we look above, we see such a relationship. Recall, we had let x = sinθ. That means that θ = sin^-1(x) or arcsin(x).

 

     Thus, we have that .

 

 

Example 2: Integrate .

 

     We know how to solve 1  x², but not 4  x². What happens if we let x = 2sinθ?

 

     Then 4  x² = 4  (2sinθ)² = 4  4sin²θ = 4(1  sin²θ) = 4cos²θ = (2cosθ)².

 

     So, in general, if we have , we should substitute x = asinθ.

 

 

     If we let x = 2sinθ, then dx = 2cosθdθ.

 

     After substituting, we have the following:

 

      

 

     But the original integral  asked us to find the indefinite integral as a function of x not θ.  What is tanθ in terms of z?  Since we substituted z = 2sinθ,

2

x

then we have .

 

q

     From here, we draw a triangle to solve for tanθ.

 

     And so, we see that .

     Thus, .

 

     Now, what happens if we have an integral of the form a² + x²?

 

     Using the second trigonometric identity we developed above, we can let x = atanθ. Then dx = asec²θdθ. (Recall, we have the relationship tan²θ + 1 = sec²θ.)

 

     Often times, though, we will need to complete the square to use a trigonometric substitution. Consider the following example:

 

 

Example 3: Evaluate .

 

     First we complete the square in the denominator by noticing that 13 = 4 + 9 and then that gives us x² + 4x + 4 + 9 = (x + 2)².

 

     .

 

     Then we let x + 2 = 3tanθ. Notice that dx = 3sec²θdθ.

 

     After substituting, we have

 

     .

 

     Solving for θ, we have .

 

     Therefore, .