Evaluate: \(\displaystyle\int{\frac{{{1}}}{{{\sin{{x}}}{\cos{{x}}}}}}{\left.{d}{x}\right.}\)

Makayla Stevens

Makayla Stevens

Answered question

2022-04-05

Evaluate:
1sinxcosxdx

Answer & Explanation

Alannah Farmer

Alannah Farmer

Beginner2022-04-06Added 11 answers

If I take the derivative of your second answer (call it g(x)), I get
dgdx=sinxcosx+sinx2(1cosx)+sinx2(1+cosx)
=sinx(1cos2x+12cosx(1+cosx)12cosx(1cosx)}{cosx(1cosx)(1+cosx)}
=sinx(1cos2x+12cosx+12cos2x12cosx+12cos2x)cosx(1cos2x)
=sinxcosxsin2x
=1cosxsinx

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