How do you find the antiderivative of #int sin(tanx)/cos^2x dx#?

Answer 1

Ezra Adams

The derivative of #tanx# is #sec^2 x# which is #1/cos^2x#. Use substitution.

#int sin(tanx)/cos^2x dx = int sin(tanx) (sec^2x) dx#

With #u = tanx#, this is #int sinu du = -cosu +C#.

So we get

#int sin(tanx)/cos^2x dx = -cos(tanx) +C#

(Check the answer by differentiation.)

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Answer 2

Emilia Aguilar

To find the antiderivative of ∫sin(tan(x))/cos²(x) dx, you can use the substitution method. Let u = tan(x), then du = sec²(x) dx. Substituting u and du into the integral yields ∫sin(u) du. Integrating sin(u) with respect to u gives -cos(u) + C. Finally, substituting back u = tan(x), the antiderivative is -cos(tan(x)) + C.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.