TRIGONOMETRIC INTEGRALS TRIGONOMETRIC SUBSTITUTIONS

This instruction material adopted of Calculus by Frank Ayres Jr 22

14. TRIGONOMETRIC INTEGRALS

THE FOLLOWING IDENTITIES are employed to find the trigonometric integrals of this chapter. 1. 1 x cos x sin 2 2 2. x sec x tan 1 2 2 3. x csc x cot 1 2 2 4. x 2 cos 1 x sin 2 1 2 5. x 2 cos 1 x cos 2 1 2 6. x 2 sin x cos x sin 2 1 7. y x sin y x sin y cos x sin 2 1 8. y x cos y x cos y sin x sin 2 1 9. y x cos y x cos y cos x cos 2 1 10. x sin 2 x cos 1 2 1 2 11. x cos 2 x cos 1 2 1 2 12. x cos 1 x sin 1 2 1 SOLVED PROBLEMS SINES AND COSINES 1. C x x dx x dx x 2 sin 2 cos 1 sin 4 1 2 1 2 1 2 2. C x x dx x dx x 6 sin 6 cos 1 3 cos 12 1 2 1 2 1 2 3. C x x dx x x dx x x dx x 3 8 1 2 2 3 cos cos sin cos 1 sin sin sin 4. dx x x dx x x dx x cos sin 1 cos cos cos 2 2 4 3 dx x x dx x x dx x cos sin cos sin 2 cos 4 2 C x x x 3 5 1 3 3 2 sin sin sin 5. dx x x x dx x x x dx x x cos sin 1 sin cos cos sin cos sin 2 2 2 2 3 2 C x x dx x x dx x x 5 5 1 3 3 1 4 2 sin sin cos sin cos sin This instruction material adopted of Calculus by Frank Ayres Jr 23

15. TRIGONOMETRIC SUBSTITUTIONS

AN INTEGRAND, which contains one of the forms 2 2 2 u b a , 2 2 2 u b a , or 2 2 2 a u b but no other irrational factor, may be transformed into another involving trigonometric functions of a new variable as follows: For Use To obtain 2 2 2 u b a z b a u sin z a z a cos sin 1 2 2 2 2 u b a z b a u tan z a z a sec tan 1 2 2 2 2 a u b z b a u sec z a z a tan 1 sec 2 In each case, integration yields an expression in the variable z. The corresponding expression in the original variable may be obtained by the use of a right triangle as shown in the solved problems below. SOLVED PROBLEMS 1. Find 2 2 4 x x dx Let x = 2 tan z; then dx = 2 sec 2 z dz and z x sec 2 4 2 dz z z z z dz z x x dx 2 2 2 2 2 tan sec 4 1 sec 2 tan 4 sec 2 4 C x x C z dz z z 4 4 sin 4 1 cos sin 4 1 2 2 2. Find dx x x 4 2 2 Let x = 2 sec z; then dx = 2 sec z tan z dz and z x tan 2 4 2 dz z dz z z z z dx x x 3 2 2 2 sec 4 tan sec 2 tan 2 sec 4 4 C z z z z tan sec ln 2 tan sec 2 C x x x x 4 ln 2 4 2 2 2 1 This instruction material adopted of Calculus by Frank Ayres Jr 24 3. Find dx x x 2 4 9 Let z x sin 2 3 ; then dz z dx cos 2 3 and z x cos 3 4 9 2 dz z z dz z z z dx x x sin cos 3 cos sin cos 3 4 9 2 2 3 2 3 2 dz z dz z dz z z sin 3 csc 3 sin sin 1 3 2 C z z z cos 3 cot csc ln 3 C x x x 2 2 4 9 4 9 3 ln 3

16. INTEGRATION BY PARTIAL FRACTIONS