How do I evaluate $\int tan (x) {sec}^{3} (x) d x$ $inttan (x) sec^3(x) dx$ ?

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How do I evaluate $\int tan (x) {sec}^{3} (x) d x$ $inttan (x) sec^3(x) dx$ ?

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Answer 1 · 2015-02-01T16:13:14

I can write the integral as:
$\int \frac{sin (x)}{cos (x)} \cdot \frac{1}{{cos}^{3} (x)} d x =$ $intSin(x)/cos(x)*1/cos^3(x)dx=$
$= \int \frac{sin (x)}{{cos}^{4} (x)} d x =$ $=intsin(x)/cos^4(x)dx=$
but: $d (cos (x)) = - sin (x) d x$ $d(cos(x))=-sin(x)dx$
and the integral becomes (multiplying and dividing by $- 1$ $-1$ ):
$= \int \frac{- sin (x)}{{- cos}^{4} (x)} d x =$ $=int(-sin(x))/-cos^4(x)dx=$
$= \int \frac{1}{{- cos}^{4} (x)} d (cos (x) =$ $=int1/-cos^4(x)d(cos(x)=$
$= - \int {cos}^{- 4} (x) d (cos (x)) =$ $=-intcos^(-4)(x)d(cos(x))=$
$= - \frac{{cos}^{- 3} (x)}{- 3} + c =$ $=-cos^(-3)(x)/(-3)+c=$
$= \frac{{cos}^{- 3} (x)}{3} + c$ $=cos^(-3)(x)/(3)+c$

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How do I evaluate $\int tan (x) {sec}^{3} (x) d x$ $inttan (x) sec^3(x) dx$ ?

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How do I evaluate $\int tan (x) {sec}^{3} (x) d x$ $inttan (x) sec^3(x) dx$ ?