We start with the expression in Equation ( 8.3.2): įinally, we return to x with the substitution u = x + 3. Using the identity sin ⁡ ( 2 ⁢ θ ) = 2 ⁢ sin ⁡ θ ⁢ cos ⁡ θ and using the reference triangle found in Key Idea 8.3.1(b), we haveġ 4 ⁢ sin ⁡ ( 2 ⁢ θ ) = 1 2 ⁢ u u 2 + 1 ⋅ 1 u 2 + 1 = 1 2 ⁢ u u 2 + 1. Now make the substitution u = tan ⁡ θ, d ⁡ u = sec 2 ⁡ θ ⁢ d ⁡ θ: † † margin: Note: Remember the sine and cosine double angle identities: sin ⁡ 2 ⁢ θ = 2 ⁢ sin ⁡ θ ⁢ cos ⁡ θ cos ⁡ 2 ⁢ θ = cos 2 ⁡ θ - sin 2 ⁡ θ = 2 ⁢ cos 2 ⁡ θ - 1 = 1 - 2 ⁢ sin 2 ⁡ θ They are often needed for writing your final answer in terms of x. SolutionWe start by completing the square, then make the substitution u = x + 3, followed by the trigonometric substitution of u = tan ⁡ θ: † † margin: 1 / 2 x 2 - 1 / 4 x θ Figure 8.3.2: A reference triangle for Example 8.3.3 Λ With a = 1 / 2, and x = 1 2 ⁢ sec ⁡ θ, we use the Pythagorean Theorem to determine the lengths of the sides of the reference triangle in Figure 8.3.2. We need to rewrite our answer in terms of x. Our original integral is given in terms of x, whereas our final answer, as given, is in terms of θ. These allow the integrand to be written in an. = 1 4 ⁢ ( sec ⁡ θ ⁢ tan ⁡ θ - ln ⁡ | sec ⁡ θ + tan ⁡ θ | ) + C. Some integrals involving trigonometric functions can be evaluated by using the trigonometric identities.

There are expressions when the earlier methods weve learned about. We integrated sec 3 ⁡ θ in Example 8.2.6, finding its antiderivatives to be The trigonometric substitution method is an important technique for integral calculus. We now rewrite the integral with these substitutions: So we have a = 1 / 2, and following Key Idea 8.3.1(c), we set x = 1 2 ⁢ sec ⁡ θ, and hence d ⁡ x = 1 2 ⁢ sec ⁡ θ ⁢ tan ⁡ θ ⁢ d ⁡ θ.

SolutionWe start by rewriting the integrand so that it has the form x 2 - a 2 for some value of a:

Trig substitution integrals