determine the following integrals: [1/(2cos^2x)-tanx cot x +(sin 2x/ cos x)] dx

Note it is necessary to simplify the integrand first
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integral = I = I (1/2cos^2x) - tanxcotx + sin2x/cosx)dx=

 I1/2cos^2x - sinx/cosx*cosx/sinx + 2sinxcosx/cosx )dx =

 I (sec^2x/2 - 1 +2sinx)dx = tanx/2 - x  - 2cosx + C

1/cos^2x = sec^2x,   I sec2xdx  = tanx

Idx = x  and  I sinxdx = - cosx
by Level 8 User (36.8k points)

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