How do you evaluate the integral #int tan^3theta#?

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Answer 1 · 2017-01-21T02:08:58

#int tan^3x dx = 1/2tan^2x +ln abs(cosx) + C#

Use the trigonometric identity:

#tan^2x = sec^2x-1#

to get:

#int tan^3x dx = int (sec^2x -1) tanx dx= int tanxsec^2xdx -int tanx dx#

Solve the first integral using: #d(tanx) = sec^2xdx#

#int tanx sec^2 dx = int tanx d(tanx) = 1/2tan^2x +C_1#

For the second integral:

#int tanx dx = int sinx/cosx dx = - int (d(cosx))/cosx = -ln abs(cosx) + C_2#

Putting it together:

#int tan^3x dx = 1/2tan^2x +ln abs(cosx) + C#