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How do you find the integral of #int sin x * tan x dx#?

Answer 1

Charles Anctil

The answer is #=ln(|tanx+secx|)-sinx+C#

We need

#tanx=sinx/cosx#

#intsecxdx=ln(tanx+secx)+C#

Consequently,

#intsinxtanxdx=intsecxsin^2xdx=intsecx(1-cos^2x)dx#

#=int(secx-cosx)dx#

#=intsecxdx-intcosxdx#

#=ln(|tanx+secx|)-sinx+C#

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Answer 2

Ezekiel Alexander

To find the integral of ∫ sin(x) * tan(x) dx, you can use substitution method. Let u = sin(x), then du = cos(x) dx. Rewrite the integral in terms of u: ∫ u * (u/√(1-u^2)) du. Now, use substitution v = 1 - u^2, then dv = -2u du. Rewrite the integral again in terms of v: ∫ -1/2 * dv. Integrating this yields -1/2 * v + C, where C is the constant of integration. Finally, substitute back for u and v to get the final answer: -1/2 * (1 - sin^2(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.