math1001-Trigonometric Integrals.pdf - MEMORIAL UNIVERSITY OF NEWFOUNDLAND DEPARTMENT OF MATHEMATICS STATISTICS Tarun Sheel MATHEMATICS 1001(Calculus II

math1001-Trigonometric Integrals.pdf - MEMORIAL UNIVERSITY...

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MEMORIAL UNIVERSITY OF NEWFOUNDLAND DEPARTMENT OF MATHEMATICS & STATISTICS Tarun Sheel MATHEMATICS 1001 (Calculus II) — Winter 2017 Unit 0: An tideriva tive (Ref. § 4.9) 1 Unit I: In def i nite In te grals 2 Unit II: Def i nite In te grals 3 Unit III: Techniques of Integration 3.1 In te gra tion by Parts (Ref. § 7.1) 3.2 The Method of Par tial Frac tions (Ref. § 7.4) 3.3 Trigonometric and Hyperbolic Integrals (Ref. § 7.2) We know that 1. Z sin( x ) dx = - cos( x ) + C 2. Z cos( x ) dx = sin( x ) + C 3. Z sec 2 ( x ) dx = tan( x ) + C 4. Z csc 2 ( x ) dx = - cot( x ) + C 5. Z sec( x ) tan( x ) dx = sec( x ) + C 6. Z csc( x ) cot( x ) dx = - csc( x ) + C 7. Z tan( x ) dx = ln | sec( x ) | + C 1
We also have 8. Z cot( x ) dx = Z cos( x ) sin( x ) dx Let u = sin( x ) ; du = cos( x ) dx = Z 1 u du = ln | u | + C = ln | sin( x ) | + C i.e. Z cot( x ) dx = ln | sin( x ) | + C 9. Z sec( x ) dx = Z sec( x ) · sec( x ) + tan( x ) sec( x ) + tan( x ) dx = Z sec 2 ( x ) + sec( x ) tan( x ) sec( x ) + tan( x ) dx Let u = sec( x ) + tan( x ) ; du = (sec( x ) tan( x ) + sec 2 ( x )) dx = Z 1 u du = ln | u | + C = ln | sec( x ) + tan( x ) | + C i.e. Z sec( x ) dx = ln | sec( x ) + tan( x ) | + C 10. Z csc( x ) dx = Z csc( x ) · csc( x ) - cot( x ) csc( x ) + cot( x ) dx = Z csc 2 ( x ) - csc( x ) cot( x ) csc( x ) + cot( x ) dx Let u = csc( x ) - cot( x ) ; du = ( - csc( x ) cot( x ) + csc 2 ( x )) dx = Z 1 u du = ln | u | + C = ln | csc( x ) - cot( x ) | + C i.e. Z csc( x ) dx = ln | csc( x ) - cot( x ) | + C You must know these, and be able to recognize corresponding substitution results; e.g Z sin( u ( x )) · u 0 ( x ) dx = - cos( u ( x )) + C 2
Examples: 1. Find Z x sec( x 2 ) dx Let u = x 2 , du = 2 xdx Z x sec( x 2 ) dx = 1 2 Z sec( u ) du = 1 2 ln | sec( u ) + tan( u ) | + C = 1 2 ln | sec( x 2 ) + tan( x 2 ) | + C 2. Find Z tan( x ) x dx Let u = x , du = 1 2 x dx Z tan( x ) x dx = 2 Z tan( u ) du = 2 ln | sec( u ) | + C = 2 ln | sec( x ) | + C We now see how we can integrate expression of the form Z sin m ( x ) cos n ( x ) dx and Z tan m ( x ) sec n ( x ) dx , where we may have m = 0 or n = 0. We can always integrate Z sin m ( x ) cos n ( x ) dx , by considering following two cases: Casae 1: m ’ and/or ’ n ’ is ODD: If power of ’sin( x )’ is ODD, keep 1 factor of ’sin’ and use Pythagorean identity to change sin 2 ( x ) = 1 - cos 2 ( x ) (a) If power of ’cos( x )’ is ODD, DO the same as above for ’cos’ (b) If both ODD, pick one that will give power ’2’ (if possible) Case 2: m ’ and ’ n ’ is EVEN: If both powers are EVEN, use Half-Angle formula sin 2 ( x ) = 1 2 (1 - cos(2 x )); OR cos 2 ( x ) = 1 2 (1 + cos(2 x )) Case 1: m and/or n a positive odd integer. We simply factor out one of the odd powers, and express the remaining even power in terms of the other function using sin 2 ( x ) = 1 - cos 2 ( x ) or cos 2 ( x ) = 1 - sin 2 ( x ), then let u be the other function; i.e. 1. Z sin 2 k +1 ( x ) cos n ( x ) dx = Z sin 2 k ( x ) · sin( x ) cos n ( x ) dx = Z (sin 2 ( x )) k cos n ( x ) sin( x ) dx = Z (1 - cos 2 ( x )) k cos n ( x ) sin( x ) dx [ u = cos( x ) , du = - sin( x ) dx ] = - Z (1 - u 2 ) k u n du etc 3
2. Z sin m ( x ) cos 2 k +1 ( x ) dx = Z sin m ( x ) · cos 2 k ( x ) cos( x ) dx = Z sin m ( x )(cos 2 ( x )) k cos( x ) dx = Z sin m ( x )(1 - sin 2 ( x )) k cos( x ) dx [ u = sin( x ) , du = cos( x ) dx ] = Z u m (1 - u 2 ) k du etc Examples: Lets find the following integrals for Case 1 : 1. Z cos 3 (2 x ) sin 5 (2 x ) dx Z cos 3 (2 x ) sin 5 (2 x ) dx = Z cos 2 (2 x ) sin 5 (2 x ) · cos(2 x ) dx = Z (1 - sin 2 (2 x )) sin 5 (2 x ) · cos(2 x ) dx Let u = sin(2 x ) ; du = 2 sin(2 x ) dx = Z (1 - u 2 ) u 5 · du 2 = 1 2 Z ( u 5 - u 7 ) du = 1 2 u 6 6 - u 8 8 = sin 6 (2 x ) 12 - sin 8 (2 x ) 16 + C 2.

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