How do you evaluate the integral $\int \frac{{tan}^{2} x}{sec x}$ $int tan^2x/secx$ ?

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How do you evaluate the integral $\int \frac{{tan}^{2} x}{sec x}$ $int tan^2x/secx$ ?

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Answer 1 · 2017-02-02T22:59:53

$ln (| sec x + tan x |) - sin x + C$ $ln(abs(secx+tanx))-sinx+C$

Explanation:

We see that:

$\int \frac{{tan}^{2} x}{sec x} d x = \int \frac{\frac{{sin}^{2} x}{{cos}^{2} x}}{\frac{1}{cos x}} d x = \int \frac{{sin}^{2} x}{cos x} d x = \int \frac{1 - {cos}^{2} x}{cos x} d x$ $inttan^2x/secxdx=int(sin^2x/cos^2x)/(1/cosx)dx=intsin^2x/cosxdx=int(1-cos^2x)/cosxdx$

Splitting up the integral:

$= \int sec x d x - \int cos x d x$ $=intsecxdx-intcosxdx$

Both of these are standard integrals:

$= ln (| sec x + tan x |) - sin x + C$ $=ln(abs(secx+tanx))-sinx+C$

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How do you evaluate the integral $\int \frac{{tan}^{2} x}{sec x}$ $int tan^2x/secx$ ?

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