Asked by Ranchit
1÷(tanx+cotx)(tanx+cotx)dx
Answers
Answered by
Steve
I assume you want the integral, but it's not clear just what you are working with.
(tanx+cotx)(tanx+cotx)
= tan^2x + 2 + cot^2x
= tan^2x+1 + cot^2x+1
= sec^2x + csc^2x
= 1/(sin^2x cos^2x)
so 1 over all that is just
1/4 sin^2(2x) = 1/8 (1-cos(4x))
and that's easy to integrate, right?
(tanx+cotx)(tanx+cotx)
= tan^2x + 2 + cot^2x
= tan^2x+1 + cot^2x+1
= sec^2x + csc^2x
= 1/(sin^2x cos^2x)
so 1 over all that is just
1/4 sin^2(2x) = 1/8 (1-cos(4x))
and that's easy to integrate, right?
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