Trig substitution
- 1. Integration Involving Trigonometric Functions and Trigonometric Substitution Dr. Philippe B. Laval Kennesaw State University September 7, 2005 Abstract This handout describes techniques of integration involving various combinations of trigonometric functions. It also describes a technique known as trigonometric substitution. Students may want to review some basic trigonometric identities before reading further. The following trigono- metric identities will be used: • sin2 x + cos2 x = 1 • 1 + tan2 x = sec2 x • sin2 x = 1 − cos 2x 2 • cos2 x = 1 + cos 2x 2 • sin 2x = 2 sin x cos x In addition, students need to remember the following: • sin x is invertible when − π 2 ≤ x ≤ π 2 . Its inverse is denoted sin−1 x. • tan x is invertible when − π 2 < x < π 2 . Its inverse is denoted tan−1 x. • sec x is invertible when 0 ≤ x < π 2 or π ≤ x < 3π 2 . Its inverse is denoted sec−1 x. 1 Powers of Sine and Cosine Before we explain the technique, let us recall that we can integrate integrals of the form sinn x cos xdxor cosn x sin xdx where n is a positive integer, by using substitution. For example, to integrate sinn x cos xdx, we let u = sin x 1
- 2. then du = cos xdx. Therefore sinn x cos xdx = un du = un+1 n + 1 = sinn+1 x n + 1 The other integral is done similarly. The technique used here depends on whether one of the powers is odd or both are even. We summarize the techniques, then do some examples. Proposition 1 Suppose we have an integral of the form sinm x cosn xdx. 1. If n is odd, that is n = 2k + 1, then save one cosine factor, and use the identity sin2 x + cos2 x = 1 to express the remaining factors in terms of sine. Then, use the substitution u = sin x. In other words sinm x cosn xdx = sinm x cos2k+1 xdx = sinm x cos2k x cos xdx = sinm x 1 − sin2 x k cos xdx 2. If m is odd, that is m = 2k + 1, then save one sine factor, and use the identity sin2 x + cos2 x = 1 to express the remaining factors in terms of cosine. Then, use the substitution u = cos x. In other words sinm x cosn xdx = sin2k+1 x cosn xdx = sin2k x cosn x sin xdx = 1 − cos2 x k cosn x sin xdx 3. If both m and n are even, we use the half-angle identities sin2 x = 1 − cos 2x 2 cos2 x = 1 + cos 2x 2 as well as the identity sin x cos x = sin 2x 2 2
- 3. We illustrate the following techniques with some examples. Example 2 Find cos3 xdx This is the case where the power of cosine is odd. We save one cosine factor and write cos3 x = cos2 x cos x = 1 − sin2 x cos x Therefore, cos3 xdx = 1 − sin2 x cos xdx = cos xdx − sin2 x cos xdx (1) The first integral is known cos xdx = sin x (2) The second integral can be evaluated using the substitution u = sin x =⇒ du = cos xdx and therefore sin2 x cos xdx = u2 du (3) = u3 3 = sin3 x 3 Using equation 2 and equation 3 in equation 1 gives us cos3 xdx = sin x − sin3 x 3 + C Example 3 Find sin5 x cos2 xdx This is the case where the power of sine is odd. We save one sine factor and write sin5 x cos2 x = sin4 x cos2 x sin x = sin2 x 2 cos2 x sin x = 1 − cos2 x 2 cos2 x sin x = 1 − 2 cos2 x + cos4 x cos2 x sin x = cos2 x sin x − 2 cos4 x sin x + cos6 x sin x 3
- 4. Therefore, sin5 x cos2 xdx = cos2 x − 2 cos4 x + cos6 x sin xdx We then use the substitution u = cos x =⇒ du = − sin xdx to get sin5 x cos2 xdx = − u2 − 2u4 + u6 du = − u3 3 − 2u5 5 + u7 7 + C = − cos3 x 3 + 2 cos5 x 5 − cos7 x 7 + C Example 4 Find sin2 xdx This is the case when the powers of sine and cosine are even (the power of cosine being 0). We use the half angle identity sin2 x = 1 − cos 2x 2 to obtain sin2 xdx = 1 2 (1 − cos 2x) dx We use the substitution u = 2x =⇒ du = 2dx to get sin2 xdx = 1 4 (1 − cos u) du = 1 4 (u − sin u) + C = 1 4 (2x − sin 2x) + C = x 2 − sin 2x 4 + C Similar techniques can be applied to powers of tangent and secant. We will not cover them here. They can be found in most Calculus books. 2 Trigonometric Substitution The techniques we are about to describe apply to integrals containing expres- sions of the form a2 − x2 a2 + x2 x2 − a2 for which the other techniques have failed. For example, if we were given x √ 1 − x2dx, the substitution u = 1 − x2 would work. However, if we were 4
- 5. given √ 1 − x2dx, it would be much more difficult to do. We will look at each case separately. Before we do this, it is important to keep in mind an important difference between the substitution technique learned before and the one we are about to explain. In the traditional substitution, we define the new variable in terms of the old. For example, u = 1 − x2 . In trigonometric substitution, we redefine the given variable. Remark 5 In order to be able to do this substitution successfully, you must be able to find all the trigonometric functions, knowing one of them. This can be done either by using trigonometric identities or a triangle. This technique can be found in any book dealing with trigonometric functions. It can also be found on the handout linked to on the web site for the class. 2.1 Integral Containing √ a2 − x2 We use the substitution x = a sin θ, with − π 2 ≤ θ ≤ π 2 and a > 0. We impose this restriction on θ so that sin θ will have an inverse. This substitution is based on the identity 1 − sin2 θ = cos2 θ and works as follows: x = a sin θ =⇒ x2 = a2 sin2 θ =⇒ a2 − x2 = a2 − a2 sin2 θ = a2 1 − sin2 θ = a2 cos2 θ Therefore a2 − x2 = √ a2 cos2 θ = |a cos θ| = |a| |cos θ| = a cos θ We were able to remove the absolute value because a > 0 and cos θ ≥ 0 when − π 2 ≤ θ ≤ π 2 . We illustrate this with examples. Example 6 Find √ 9 − x2 x2 dx We let x = 3 sin θ, with − π 2 ≤ θ ≤ π 2 . Then dx = 3 cos θ. Also, as noted above, 5
- 6. √ 9 − x2 = 3 cos θ. Therefore, √ 9 − x2 x2 dx = 3 cos θ 9 sin2 θ 3 cos θdθ = cos2 θ sin2 θ dθ = cot2 θdθ = csc2 θ − 1 dθ = − cot θ − θ + C We need to express our answer in terms of x. Since x = 3 sin θ, it follows that θ = sin−1 x 3 . Also, either using trigonometric identities, or a triangle, we find that cot θ = √ 9 − x2 x . Therefore, √ 9 − x2 x2 dx = − √ 9 − x2 x − sin−1 x 3 + C Example 7 Find 2 0 √ 4 − x2dx Method 1 We recognize that √ 4 − x2 is the upper half circle of radius 2 cen- tered at the origin. The integral of it between 0 and 2 corresponds to the area of the first quadrant of this circle. Therefore 2 0 4 − x2dx = 1 4 (area of a circle of radius 2) = 1 4 22 π = π This method is very quick and easy. However, it would not work if the problem had been to find an antiderivative. We show another technique, using trigonometric substitution. Method 2 According to what was explained above, we let x = 2 sin θ. Then,√ 4 − x2 = 2 cos θ. Also, dx = 2 cos θdθ. To find the value of this integral, we will first find an antiderivative, then use the given limits of integration. Therefore, 4 − x2dx = 4 cos θ cos θdθ = 4 cos2 θdθ 6
- 7. Remembering the techniques of the previous section, we use cos2 θ = 1 + cos 2θ 2 . Therefore, 4 − x2dx = 4 1 + cos 2θ 2 dθ = 2 (1 + cos 2θ) dθ If we let u = 2θ, then du = 2dθ and we have 4 − x2dx = (1 + cos u) du = u + sin u = 2θ + sin 2θ = 2θ + 2 sin θ cos θ We obtained the last equality using the identity sin 2θ = 2 sin θ cos θ. Now, we write everything back in terms of x. First, since x = 2 sin θ, we see that sin θ = x 2 and θ = sin−1 x 2 To express cos θ in terms of x, we use cos2 θ = 1−sin2 θ and since cos θ ≥ 0, we have cos θ = 1 − sin2 θ = 1 − x2 4 = 4 4 − x2 4 = 1 4 (4 − x2) = √ 4 − x2 2 Therefore 4 − x2dx = 2 sin−1 x 2 + 2 x 2 √ 4 − x2 2 = 2 sin−1 x 2 + x √ 4 − x2 2 7
- 8. We can now find the definite integral 2 0 4 − x2dx = 2 sin−1 x 2 + x √ 4 − x2 2 2 0 = 2 sin−1 1 + 2 √ 0 2 − 2 sin−1 0 + 0 √ 4 2 = 2 π 2 + 0 − (0 + 0) = π 2.2 Integral Containing √ a2 + x2 We use the substitution x = a tan θ, with a > 0 and − π 2 < θ < π 2 . We impose this restriction on θ so that tan θ will have an inverse. The substitution is based on the identity 1 + tan2 θ = sec2 θ and works as follows: a2 + x2 = a2 + a2 tan2 θ = a2 1 + tan2 θ = a2 sec2 θ Therefore a2 + x2 = √ a2 sec2 θ = a √ sec2 θ = a |sec θ| = a sec θ Because a > 0 and sec θ ≥ 0 if − π 2 < θ < π 2 . Example 8 Find 1 x2 √ x2 + 4 dx We let x = 2 tan θ, dx = 2 sec2 θdθ. Also, √ x2 + 4 = 2 sec θ. Therefore: 1 x2 √ x2 + 4 dx = 2 sec2 θdθ 4 tan2 θ (2 sec θ) = 1 4 sec θ tan2 θ dθ Now, sec θ tan2 θ = 1 cos θ sin2 θ cos2 θ = cos θ sin2 θ 8
- 9. If we make the substitution u = sin θ, then du = cos θdθ and we get: 1 x2 √ x2 + 4 dx = 1 4 cos θ sin2 θ dθ = 1 4 du u2 = 1 4 u−2 du = −1 4u + C = −1 4 sin θ + C We express sin θ in terms of x and obtain sin θ = x √ 4 + x2 Therefore 1 x2 √ x2 + 4 dx = − √ 4 + x2 4x + C 2.3 Integral Containing √ x2 − a2 We use the substitution x = a sec θ, with a > 0 and 0 ≤ θ < π 2 or π ≤ θ < 3π 2 . We impose this restriction on θ so that sec θ will be invertible. This substitution is based on the identity sec2 θ − 1 = tan2 θ and works as follows: x2 − a2 = a2 sec2 θ − a2 = a2 sec2 θ − 1 = a2 tan2 θ Therefore x2 − a2 = a2 tan2 θ = |a| |tan θ| = a tan θ because a > 0 and tan θ ≥ o when 0 ≤ θ < π 2 or π ≤ θ < 3π 2 . Example 9 Find dx √ x2 − a2 , where a > 0. According to the explanation above, we let x = a sec θ. Then, dx = a sec θ tan θdθ. 9
- 10. Also, √ x2 − a2 = a tan θ. Therefore dx √ x2 − a2 = a sec θ tan θdθ a tan θ = sec θdθ = ln |sec θ + tan θ| + C (see homework 1). Now, we need to write everything in terms of x. sec θ = x a and tan θ = √ x2 − a2 a . Therefore, dx √ x2 − a2 = ln x a + √ x2 − a2 a + C = ln x + x2 − a2 − ln a + C = ln x + x2 − a2 + C 3 Problems 1. Find sec θdθ. (hint: multiply both numerator and denominator by sec θ + tan θ) 2. Find cos5 x sin5 xdx 3. Find cos4 xdx 4. Using the technique of example 7, find √ a2 − x2dx 5. Find x √ 1 − x2 dx 6. Find √ 1 − 4x2dx 7. Find x √ x2 + 3 dx 8. Find et √ 9 − e2tdt 4 Answers 1. sec θdθ = ln |sec θ + tan θ| + C 2. cos5 x sin5 xdx = 1 6 sin6 x − 1 4 sin8 x + 1 10 sin10 x + C 3. cos4 xdx = 3 8 x + 1 4 sin 2x + 1 32 sin 4x 10
- 11. 4. √ a2 − x2dx = x 2 √ a2 − x2 + a2 2 sin−1 x a + C 5. x √ 1 − x2 dx = − √ 1 − x2 + C 6. √ 1 − 4x2dx = 1 4 sin−1 2x + 1 2 x √ 1 − 4x2 + C 7. x √ x2 + 3 dx = √ x2 + 3 + C 8. et √ 9 − e2tdt = 1 2 et √ 9 − e2t + 9 sin−1 et 3 + C 11