Lecture 2 - Trigonometric Integrals.pdf - 2 – Trigonometric Integrals We know the antiderivatives of sine and cosine � sin(x)dx = − cos(x C �

Lecture 2 - Trigonometric Integrals.pdf - 2 –...

This preview shows page 1 out of 8 pages.

Unformatted text preview: 2 – Trigonometric Integrals We know the antiderivatives of sine and cosine: ∫ sin(x)dx = − cos(x) + C ∫ cos(x)dx = sin(x) + C Using substitution ( u = cos( x ) ) we can show, ∫ tan(x)dx = ∫ sin(x) 1 dx = − ∫ du = −ln | cos(x) | +C = ln sec(x) + C cos(x) u Let’s evaluate some more complicated integrals involving trigonometric functions. Consider integrals of the form: m n sin ( x ) cos (x)dx ∫ (where m and n are non-negative integers) 1 m n sin ( x ) cos (x)dx Integrals of the form ∫ First consider sin(x) cos(x)dx ∫ We can evaluate this with the substitution u = sin( x ) du = cos( x )dx 1 2 1 2 ∫ sin( x) cos(x)dx = ∫ u du = 2 u + C = 2 sin (x) + C Notice, we could have instead used the substitution u = cos( x ) du = − sin( x )dx 1 2 ~ 1 2 ~ ∫ sin( x) cos(x)dx = −∫ u du = − 2 u + C = − 2 cos (x) + C2 So we have 1 2 ∫ sin( x) cos(x)dx = 2 sin (x) + C and 1 2 ~ ∫ sin( x) cos(x)dx = − 2 cos (x) + C These look different but remember the Pythagorean Identity 2 2 sin (x) + cos (x) = 1 1 2 1 ~ ~ 1 2 ~ 1 1 2 2 − cos ( x ) + C = − (1 − sin ( x )) + C = sin ( x ) + (C − ) = sin ( x ) + C 2 2 2 2 2 The two families of functions are equivalent due to the 3 arbitrary integration constant. Integrals with powers of sine can be evaluated easily if there is a single cosine present (and vice versa): That is, with u = sin( x ) du = cos( x )dx , we have 1 m+1 1 m +1 ∫ sin (x) cos(x)dx = ∫ u du = m + 1 u + C = m + 1 sin (x) + C m m Similarly, with u = cos( x ) du = − sin( x )dx 1 n +1 1 n +1 sin( x ) cos ( x ) dx = − u du = − u + C = − cos (x) + C ∫ ∫ n +1 n +1 n n But what if neither sine nor cosine appear with a power of one? 4 Example: Sine and/or cosine has an odd power. We make use of the Pythagorean Trig Identity again to reduce the sine or cosine with an odd power until just a single power remains and then use the previous substitutions. For example, ∫ sin (x) cos (x)dx = ∫ sin (x)(1− sin (x))cos(x)dx = ∫ (sin (x) − sin (x))cos(x)dx u = sin( x ) = ∫ (u − u )du du = cos( x )dx 2 3 2 2 2 2 4 4 1 3 1 5 = sin ( x ) − sin ( x ) + C 3 5 5 Example: Sine and cosine both have even power. In this case, we make use of the half-angle identities, 1 − cos( 2x ) sin ( x ) = 2 1 + cos( 2x ) cos ( x ) = 2 2 2 This turns the integrand into a polynomial with even powers of cos( 2x ) so half-angle identities may need to be applied recursively to complete the integration. 1 2 2 E.g., ∫ sin ( x ) cos ( x )dx = 2 ∫ (1 − cos( 2x ) )(1 + cos( 2x ) )dx 2 1 2 = ∫ 1 − cos ( 2x ) dx 4 1 ⎡ 1 ⎤ Exercise: Complete = ∫ ⎢1 − (1 + cos( 4x ) )⎥ dx the calculation. 4 ⎣ 2 6 ⎦ [ ∫ tan Integrals of the form m n (x) sec (x)dx can be treated similarly using the Pythagorean identity 2 2 sec (x) = tan (x) + 1 and the substitution u = tan( x ) Z e.g. Z tan3 (x) sec2 (x) dx = Z <latexit sha1_base64="rdPVxjDp5fEafYHDCrf4gwUEZw4=">AAACGHicbVDLSsNAFJ3UV62vqEs3g0WoIDVpBd0IRTcuK9gHNGmZTCbt0MkkzEykJfQz3Pgrblwo4rY7/8Zp2oVWDwwczjmXufd4MaNSWdaXkVtZXVvfyG8WtrZ3dvfM/YOmjBKBSQNHLBJtD0nCKCcNRRUj7VgQFHqMtLzh7cxvPRIhacQf1Dgmboj6nAYUI6WlnnnuUK6goxDvVkujU+hIgruVjJ1BfwSvYRZIutVMSHpm0SpbGeBfYi9IESxQ75lTx49wEhKuMENSdmwrVm6KhKKYkUnBSSSJER6iPuloylFIpJtmh03giVZ8GERCP71Fpv6cSFEo5Tj0dDJEaiCXvZn4n9dJVHDlppTHiSIczz8KEgZVBGctQZ8KghUba4KwoHpXiAdIIKx0lwVdgr188l/SrJRtq2zfXxRrN4s68uAIHIMSsMElqIE7UAcNgMETeAFv4N14Nl6ND+NzHs0Zi5lD8AvG9Bt2ipzg</latexit> tan2 (x) sec4 (x) dx = = = <latexit sha1_base64="ETX/xpBiqfBwdjC0CDjI5LRJNOk=">AAACh3icfZFdT8IwFIa7iQr4hXrpTSPR4EdwQ6LcmKDeeImJfCQMSFc6aOi6pe0MhPBX/FHe+W/sBjEKxJO0eXLe96Sn57gho1JZ1pdhbqQ2t7bTmezO7t7+Qe7wqCGDSGBSxwELRMtFkjDKSV1RxUgrFAT5LiNNd/Qc6813IiQN+JuahKTjowGnHsVI6VQv9+FQrqCjEO+WCuML6EiCu+WErmF/DM8f4DrHMiVex8mu2gs/eGX/WxJ1S7Cgr8SmxSgWe7m8VbSSgKtgLyAPFlHr5T6dfoAjn3CFGZKybVuh6kyRUBQzMss6kSQhwiM0IG2NHPlEdqbJHGfwTGf60AuEPrqjJPu7Yop8KSe+q50+UkO5rMXJdVo7Ul6lM6U8jBTheP6QFzGoAhgvBfapIFixiQaEBdW9QjxEAmGlVxcPwV7+8io0SkXbKtqv5Xz1aTGONDgBp6AAbHAPquAF1EAdYCNlXBq3RtnMmDfmnVmZW01jUXMM/oT5+A1h6ree</latexit> Z Z Z du = sec2 ( x )dx u3 du tan2 (x) sec2 (x) sec2 (x) dx tan2 (x)(tan2 (x) + 1) sec2 (x) dx u2 (u2 + 1) du 7 Another useful substitution u = sec( x ) du = tan( x ) sec( x )dx Z Z sec3 (x) tan(x) dx = sec2 (x) sec(x) tan(x) dx Z = u2 du <latexit sha1_base64="eY02kuKkfpJjix3x/K+pWop9Xpc=">AAACRnicbVDPS8MwGP06f835a+rRS3AoCjLaKehFGHrxOMFtwlpHmmUaTNOSpOIo++u8ePbmn+DFgyJeTbuKuvkg5PHe+8iX50ecKW3bz1Zhanpmdq44X1pYXFpeKa+utVQYS0KbJOShvPSxopwJ2tRMc3oZSYoDn9O2f3ua+u07KhULxYUeRNQL8LVgfUawNlK37LlMaOQqSq72d+53kauxyO491LtH28fox69lumGTOdctfUfjq1omxqnYLVfsqp0BTRInJxXI0eiWn9xeSOKACk04Vqrj2JH2Eiw1I5wOS26saITJLb6mHUMFDqjykqyGIdoySg/1Q2mO2SRTf08kOFBqEPgmGWB9o8a9VPzP68S6f+QlTESxpoKMHurHHOkQpZ2iHpOUaD4wBBPJzK6I3GCJiTbNpyU441+eJK1a1bGrzvlBpX6S11GEDdiEHXDgEOpwBg1oAoEHeIE3eLcerVfrw/ocRQtWPrMOf1CALxZuquw=</latexit> We then need to use the fact that: ∫ sec(x)dx = ln sec(x) + tan(x) + C 8 ...
View Full Document

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture