What's the integral of $\int sin (x) tan (x) d x$ $int sin(x)tan(x) dx$ ?

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Answer 1 · 2015-11-21T17:19:02

$\int sin (x) tan (x) d x = ln | sec (x) + tan (x) | - sin (x) + c$ $intsin(x)tan(x)dx = ln|sec(x) + tan(x)| - sin(x) + c$

We have

$\int sin (x) tan (x) d x$ $intsin(x)tan(x)dx$

Or, in other terms

$\int \frac{{sin}^{2} (x)}{cos (x)} d x$ $intsin^2(x)/cos(x)dx$

Or, since ${sin}^{2} (x) = 1 - {cos}^{2} (x)$ $sin^2(x) = 1 - cos^2(x)$

$\int \frac{1 - {cos}^{2} (x)}{cos (x)} d x$ $int(1 - cos^2(x))/cos(x)dx$

Doing the fraction we have

$\int (sec (x) - cos (x)) d x$ $int(sec(x) - cos(x))dx$

Break the integral into fractions

$\int sec (x) d x - \int cos (x) d x$ $intsec(x)dx - intcos(x)dx$

And those are standard integrals, so evaluate that

$\int sin (x) tan (x) d x = ln | sec (x) + tan (x) | - sin (x) + c$ $intsin(x)tan(x)dx = ln|sec(x) + tan(x)| - sin(x) + c$

What's the integral of int sin(x)tan(x) dx∫sin(x)tan(x)dxint sin(x)tan(x) dx?