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4.4 Trigonometric Substitutions 4.4 Brian E. Veitch Trigonometric Substitutions Trig substitution reduces integrals of a certain form to integrals of trig functions, which can be easier to deal with. The idea is to match the integral like so If you see √ a2 − x 2 √ a2 + x 2 √ x 2 − a2 Substitute Uses the following Identity x = a sin(θ) 1 − sin2 (θ) = cos2 (θ) x = a tan(θ) 1 + tan2 (θ) = sec2 (θ) x = a sec(θ) sec2 (θ) − 1 = tan2 (θ) I think it’s best we just get right into it. Z Example 4.5. Find √ dx 25 + x2 Since the bottom has the form a2 + x2 , we use the second option. Let x = 5 tan(θ), dx = 5 sec2 (θ) dθ. So, Z dx √ = 25 + x2 Z Z 5 sec2 (θ) dθ p = 25 + 25 tan2 (θ) 5 sec2 (θ) dθ = 5 sec(θ) 5 sec2 (θ) dθ Z p 25(1 + tan2 (θ)) Z = 5 sec2 (θ) dθ p 25 sec2 (θ) Z sec(θ) dθ = ln | sec(θ) + tan(θ)| + C That didn’t seem to bad, right? Well, we’re not done yet. Remember, this is a substitution problem. The original variable is x. So we need to substitute back. Sometimes it’s easy, and sometimes you need a little trig. 137 4.4 Trigonometric Substitutions Brian E. Veitch 1. So what’s tan(θ) Since x = 5 tan(θ), we get tan(θ) = x . 5 2. And sec(θ) ? Let’s take a look at a right triangle with angle θ. √ 25 + x2 . From here, we see that sec(θ) = 5 Therefore, Z or Z √ 25 + x2 x dx √ = ln + +C 2 5 5 25 + x √ dx √ = ln | 25 + x2 + x| − ln 5 + C 25 + x2 138 4.4 Trigonometric Substitutions Z Example 4.6. How about √ Brian E. Veitch x dx 25 + x2 You need to be careful. When you’re working through a textbook, all the problems in those sections are usually done using the same method. In this case that method would be trig substitution. And it would work. But, this integral can be done using u-substitution. 1. Using Trig Substitution Let x = 5 tan(θ), dx = 5 sec2 (θ) dθ Z x dx √ = 25 + x2 Z 25 tan(θ) sec2 (θ) dθ p = 25 sec2 (θ) Z 25 tan(θ) sec2 (θ) dθ = 5 sec(θ) Z √ √ 25 + x2 5 sec(θ) + C = 5 + C = 25 + x2 + C 5 We used the right triangle from above to change the sec(θ). 2. Using u-substitution Let u = 25 + x2 , du = 2x dx → Z du = x dx 5 Z x dx 1 −1/2 √ = u du 2 25 + x2 = u1/2 + C √ = 25 + x2 + C So which one is faster? 139 5 sec(θ) tan(θ) dθ 4.4 Trigonometric Substitutions Z Example 4.7. Find Brian E. Veitch √ x3 9 − x2 Since the denominator has the form a2 − x2 , we use the substitution x = 3 sin(θ), and dx = 3 cos(θ) dθ Z Z q √ 3 2 x 9 − x dx = 27 sin (θ) 9 − 9 sin2 (θ) · (3 cos(θ)) dθ Z p = 27 sin3 (θ) 9 cos2 (θ) · 3 cos(θ) dθ Z 5 = 3 sin3 (θ) cos2 (θ) dθ Z 5 = 3 sin2 (θ) cos2 (θ) · sin(θ) dθ Z 5 = 3 (1 − cos2 (θ)) cos2 (θ) · sin(θ) dθ 3 Let u = cos(θ), and du = − sin(θ) dθ 5 3 Z 2 2 5 Z (1 − cos (θ)) cos (θ) · sin(θ) dθ = −3 5 (1 − u2 )u2 du Z u2 − u4 du 1 5 5 1 3 u − u = −3 3 5 3 cos5 (θ) 5 cos (θ) = −3 − 3 5 = −3 Now we just have to figure out what cos(θ) is. To do that, we use the right triangle again. Recall that x = 3 sin(θ) 140 4.4 Trigonometric Substitutions Brian E. Veitch √ 9 − x2 (9 − x2 )1/2 = . Now we just replace the cos(θ) with this We see that cos(θ) = 3 3 in our integral and we’re done! √ 3 (9 − x2 )3/2 9 − x2 3 = . Note cos (θ) = 3 33 Z x 3 √ 9− x2 1 1 2 3/2 2 5/2 (9 − x ) − (9 − x ) +C dx = −3 34 5 · 35 (9 − x2 )3/2 = − 15 − (9 − x2 ) + C 5 (9 − x2 )3/2 2 = − x +6 +C 5 5 The last few steps weren’t necessary, but it doesn’t hurt to practice simplifying. Z Example 4.8. Find x2 √ 1 x2 − 36 Let x = 6 sec(θ), and dx = 6 sec(θ) tan(θ)dθ. Before setting up the integral, we can substitute and simplify √ x2 − 36 = √ p 36(sec2 (θ) − 9) = 6 tan(θ) 141 x2 − 36 right now. 4.4 Trigonometric Substitutions Brian E. Veitch and x2 = 36 sec2 (θ) Now we substitute, Z Z 1 1 √ = · 6 sec(θ) tan(θ) dθ 2 2 2 36 sec (θ) · 6 tan(θ) x x − 36 Z 1 dθ = 36 sec(θ) Z 1 = cos(θ) dθ 36 1 = sin(θ) + C 36 We need to use the right triangle to rewrite sin(θ) in terms of x. If x = 6 sec(θ), then x 6 = sec(θ) → cos(θ) = 6 x √ x2 − 36 x √ Z 1 1 x2 − 36 √ Therefore, = · +C 36 x x2 x2 − 36 From here you can see sin(θ) = 142 4.4 Trigonometric Substitutions Z Example 4.9. Find √ Brian E. Veitch x2 dx 9 − x2 Since the denominator has the form a2 − x2 , we use the following substitution, x = 3 sin(θ) , dx = 3 cos(θ) To help simplify before we start the intergral, √ Z 9− x2 q = 9 − 9 sin2 (θ) = 3 cos(θ) Z x 9 sin2 (θ) √ · 3 cos(θ) dθ dx = 3 cos(θ) 9 − x2 Z = 9 sin2 (θ) dθ Z 1 = 9 (1 − cos(2θ)) dθ 2 Z 9 1 − cos(θ) dθ = 2 9 1 = θ − sin(2θ) + C 2 2 9 9 = θ − sin(2θ) + C 2 4 Now we use the right triangle, 143 4.4 Trigonometric Substitutions Brian E. Veitch Note that we can’t use sin(2θ), since the angle is just a single θ. To deal with this, we use the trig identity sin(2θ) = 2 sin(θ) cos(θ) Z 9 9 x2 √ dx = θ − sin(θ) cos(θ) 2 2 9 − x2 −1 Using the triangle, we have θ = sin x x , sin(θ) = , and cos(θ) = 3 3 Final Answer: x√ 9 − x2 9 x2 −1 x √ dx = sin − +C 2 3 9 9 − x2 Z Z Example 4.10. Find x √ dx 2 x + 4x + 5 144 √ 9 − x2 3 4.4 Trigonometric Substitutions Brian E. Veitch The problem here is what’s in the square root doesn’t follow one of our three forms, which are x2 − a2 , x2 + a2 , or a2 − x2 . So we need to complete the square on x2 + 4x + 5 so it has one of these forms. x2 + 4x + 5 = x2 + 4x+? −? + 5 The missing number ?, is ? = −b 2a 2 = −4 2(1) 2 = (−2)2 = 4. x2 + 4x + 5 = x2 + 4x + 4 − 4 + 5 x2 + 4x + 5 = (x + 2)2 + 1 Now we have one of the correct forms. It has the form x2 + a2 , well sorta. But here’s what we have now. Z x √ dx = 2 x + 4x + 5 Z We need to substitute. 1. Let u = x + 2 and x = u − 2 145 x p dx (x + 2)2 + 1 4.4 Trigonometric Substitutions Brian E. Veitch 2. du = dx 3. Substitute Z Z x p dx = (x + 2)2 + 1 u−2 √ du u2 + 1 4. Now we do the trig substitution Let x = 1 tan(θ), du = sec2 (θ) dθ. √ Z u2 q p + 1 = tan2 (θ) + 1 = sec2 (θ) = sec(θ) x u−2 √ du u2 + 1 Z tan(θ) − 2 = · sec2 (θ) dθ sec(θ) Z = sec(θ) tan(θ) − 2 sec(θ) dθ Z = sec(θ) − 2 sec(θ) Z p dx = (x + 2)2 = sec(θ) − 2 ln | sec(θ) + tan(θ)| + C Now we use the right triangle. 146 4.4 Trigonometric Substitutions √ From here we get sec(θ) = Brian E. Veitch u2 + 1 and tan(θ) = u 1 sec(θ) − 2 ln | sec(θ) + tan(θ)| + C = Z √ √ √ u2 + 1 − 2 ln u2 + 1 + u + C √ √ x = x2 + 4x + 5 − 2 ln x2 + 4x + 5 + (x + 2) + C x2 + 4x + 5 147