Asked by Janice
I'm working this problem:
∫ [1-tan^2 (x)] / [sec^2 (x)] dx
∫(1/secx)-[(sin^2x/cos^2x)/(1/cosx)
∫cosx-sinx(sinx/cosx)
∫cosx-∫sin^2(x)/cosx
sinx-∫(1-cos^2(x))/cosx
sinx-∫(1/cosx)-cosx
sinx-∫secx-∫cosx
sinx-sinx-∫secx
=-ln |secxtanx|+C
Can someone just verify that I did everything correctly?
∫ [1-tan^2 (x)] / [sec^2 (x)] dx
∫(1/secx)-[(sin^2x/cos^2x)/(1/cosx)
∫cosx-sinx(sinx/cosx)
∫cosx-∫sin^2(x)/cosx
sinx-∫(1-cos^2(x))/cosx
sinx-∫(1/cosx)-cosx
sinx-∫secx-∫cosx
sinx-sinx-∫secx
=-ln |secxtanx|+C
Can someone just verify that I did everything correctly?
Answers
Answered by
Steve
I think your 2nd line is bogus, and the brackets are unbalanced.
(1-tan^2)/sec^2
= 1/sec^2 - tan^2/sec^2
= cos^2 - sin^2
= cos(2x)
integral of cos(2x) = 1/2 sin(2x)
(1-tan^2)/sec^2
= 1/sec^2 - tan^2/sec^2
= cos^2 - sin^2
= cos(2x)
integral of cos(2x) = 1/2 sin(2x)
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