Trigonometric Integrals

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1 Most trigonometric integrals can be solved by using trigonometric identities or by following a strategy based on the form of the integrand. There are some that are not so easy! Basic Trig Identities and Integrals Pythagorean Identities. If you can remember the first of these, then () and (3) can be found by dividing both sides of (1) by cos (x) and sin (x), respectively: (1) sin (x) + cos (x) 1 () tan (x) + 1 sec (x) (3) 1 + cot (x) csc (x) Cosine Double-Angle Power-Reducing Identities. The cosine double-angle identities are a direct consequence of the cosine sum identity cos(a + B) cos(a) cos(b) sin(a) sin(b) with A B x and the Pythagorean identity (1). manipulating the cosine double-angle identities: The power-reducing identities follow by (1) cos(x) 1 sin (x) sin (x) 1 cos(x) () cos(x) cos (x) 1 cos (x) 1 + cos(x) A Useful Trig Integrals Theorem. The following six trig integrals are helpful to know: cos(x) dx sin(x) + C (1) sin(x) dx cos(x) + C () tan(x) dx ln cos(x) + C (3) cot(x) dx ln sin(x) + C (4) sec(x) dx ln sec(x) + tan(x) + C (5) csc(x) dx ln csc(x) + cot(x) + C (6) 1

2 We already know Integrals (1) and (). Integral (3) is obtained using tan(x) sin(x) cos(x) and the substitution p cos(x). Similarly, Integral (4) is the result of using cot(x) cos(x) and the sin(x) substitution p sin(x). Integral (5) results from a little trick: multiply the integrand sec(x) by sec(x) + tan(x) and then use the substitution p sec(x) + tan(x). Similarly, for Integral (6) we sec(x) + tan(x) multiply the integrand csc(x) by csc(x) + cot(x) csc(x) + cot(x) Be careful with those minus signs in Integrals (), (3), and (6)! and then use the substitution p csc(x) + cot(x). A Strategy for Solving sin m (x) cos n (x) dx We assume that m and n are non-negative integers. If the power of cosine is odd (that is, n k + 1 for some positive integer k), we pull-off one factor of cosine and use the Pythagorean identity cos (x) 1 sin (x) to express the rest of the integrand in terms of powers of sine, then use the substitution p sin(x), dp cos(x) dx: sin m (x) cos k+1 (x) dx sin m (x) ( cos (x) ) k cos(x) dx sin m (x) ( 1 sin (x) ) k[ cos(x) dx ] p m (1 p ) k dp. If the power of sine is odd (that is, m k+1 for some positive integer k), we pull-off one factor of sine and use the Pythagorean identity sin (x) 1 cos (x) to express the rest of the integrand in terms of powers of cosine, then use the substitution p cos(x), dp sin(x) dx, or equivalently, sin(x) dx dp: sin k+1 (x) cos n (x) dx ( sin (x) ) k cos n (x) sin(x) dx (1 cos (x) ) k cos n (x) [ sin(x) dx ] (1 p ) k p n dp.

3 A Strategy for Solving sin m (x) cos n (x) dx (cont.) If the powers of both cosine and sine are odd, we may use either of the above strategies. The first may be a better choice since we don t have to worry about the negative that gets introduced in the second. If the powers of both cosine and sine are even, then we use the power-reducing identities to rewrite the integrand in terms of double-angles. In general, we must continue to rewrite the integrand until all powers of cosine are 1! EXAMPLE #1. Solve sin 4 (x) dx. Solution. Note that the powers of sine and cosine are even (the power of cosine is 0!), so we apply the power-reducing identity to sin (x): ( ) 1 cos(x) sin 4 (x) (sin (x)) 1 ( 1 cos(x) + cos (x) ). 4 Now, applying the power-reducing identity to cos (x) gives cos (x) 1 + cos(4x). Thus, we have sin 4 (x) 1 4 ( 1 cos(x) cos(4x) ) cos(x) cos(4x), and all powers of cosine are 1. We may now solve the given integral using appropriate substitutions, say p x and q 4x: sin 4 (x) dx dx 1 cos(x) dx cos(4x) dx 3 8 x 1 ( ) 1 cos(p) dp ( ) 1 cos(q) 4 dq 3 8 x 1 4 sin(p) sin(q) + C 3 8 x 1 4 sin(x) + 1 sin(4x) + C. 3 3

4 EXAMPLE #. Solve sin 6 (x) cos 5 (x) dx. Solution. The power of cosine is odd, so we are going to pull-off (save for later!) one factor of cosine and use the Pythagorean identity cos (x) 1 sin (x) to rewrite the integrand: sin 6 (x) cos 5 (x) sin 6 (x) ( cos (x) ) [ cos(x) ] sin 6 (x) ( 1 sin (x) ) [ cos(x) ]. Next, we let p sin(x), so that dp cos(x) dx. This gives sin 6 (x) cos 5 (x) dx sin 6 (x) ( 1 sin (x) ) [ ] cos(x) dx p 6 (1 p ) dp p 6 (1 p + p 4 ) dp (p 6 p 8 + p 10 ) dp p7 7 ( p 9 9 ) + p C 1 7 sin7 (x) 9 sin9 (x) sin11 (x) + C. EXAMPLE #3. Show that sin 5 (x) cos (x) dx 1 3 cos3 (x) + 5 cos5 (x) 1 7 cos7 (x) + C. 4

5 A Strategy for Solving tan m (x) sec n (x) dx We assume that m and n are non-negative integers. If the power of tangent is odd (that is, m k + 1 for some positive integer k), we pull-off a factor of sec(x) tan(x) and use the Pythagorean identity tan (x) sec (x) 1 to express the rest of the integrand in terms of powers of secant, then use the substitution p sec(x), dp sec(x) tan(x) dx: tan k+1 (x) sec n (x) dx ( tan (x) ) k sec n 1 (x) sec(x) tan(x) dx ( sec (x) 1 ) k sec n 1 (x) [ sec(x) tan(x) dx ] (p 1) k p n 1 dp. If the power of secant is even but not zero (that is, n k for some positive integer k), we pull-off one factor of sec (x) and use the Pythagorean identity sec (x) 1 + tan (x) to express the rest of the integrand in terms of powers of tangent, then use the substitution p tan(x), dp sec (x) dx: tan m (x) sec k (x) dx tan m (x) ( sec (x) ) k 1 sec (x) dx tan m (x) ( 1 + tan (x) ) k 1[ sec (x) dx ] p m (1 + p ) k 1 dp. For all other cases, there is no official strategy. We may need to use identities, integration by parts, and sometimes we must be very clever, that is, use some kind of trick. Also, Integrals (3) and (5) given on page 1 may be needed. 5

6 EXAMPLE #4. Solve tan 7 (x) sec 9 (x) dx. Solution. The power of tangent is odd, so we will pull-off a factor of sec(x) tan(x) and use the Pythagorean identity tan (x) sec (x) 1 to rewrite the integrand: tan 7 (x) sec 9 (x) ( tan (x) ) 3 sec 8 (x) [ sec(x) tan(x) ] ( sec (x) 1 ) 3 sec 8 (x) [ sec(x) tan(x) ]. Now, let p sec(x), so that dp sec(x) tan(x) dx. Then our given integral can be rewritten as follows: tan 7 (x) sec 9 (x) dx ( sec (x) 1 ) 3 sec 8 (x) [ sec(x) tan(x) ] dx (p 1 ) 3 p 8 dp (p 6 3p 4 + 3p 1) p 8 dp (p 14 3p 1 + 3p 10 p 8 ) dp ( ) ( ) p15 p 13 p p C 1 15 sec15 (x) 3 13 sec13 (x) sec11 (x) 1 9 sec9 (x) + C. EXAMPLE #5. Solve tan 6 (x) sec 4 (x) dx. Solution. The power of secant is even, so we will pull-off a factor of sec (x) and use the Pythagorean identity sec (x) 1 + tan (x) to rewrite the integrand: tan 6 (x) sec 4 (x) tan 6 (x) sec (x) [ sec (x) ] tan 6 (x) ( 1 + tan (x) )[ sec (x) ]. 6

7 Solution. (cont.) Now, let p tan(x), so that dp sec (x) dx. Then our given integral can be rewritten as follows: tan 6 (x) sec 4 (x) dx tan 6 (x) ( 1 + tan (x) )[ sec (x) dx ] p 6 (1 + p ) dp (p 6 + p 8 ) dp p7 7 + p9 9 + C 1 7 tan7 (x) tan9 (x) + C. EXAMPLE #6. Show that x sec(x) tan(x) dx x sec(x) ln sec(x) + tan(x) + C. EXAMPLE #7. (A) Prove the following reduction formula for sine: sin n (x) dx 1 n sinn 1 (x) cos(x) + n 1 n sin n (x) dx, where n is an integer. Note that sin 0 (x) 1. Hint: The form of the right-hand side looks very familiar... if u sin n 1 (x), then what is dv? (B) Use the sine reduction formula to rework EXAMPLE #1. You will need to apply the reduction formula twice. Compare the antiderivatives that you obtain. Are the two antiderivatives really the same function? (C) Prove the reduction formula for secant: sec n (x) dx 1 n 1 secn (x) tan(x) + n n 1 sec n (x) dx. 7

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