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

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

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Answer 1 · 2015-02-18T14:17:41

I would start by writing your integrand as:

$\int \frac{1}{{cos}^{3} (x)} \cdot \frac{sin (x)}{cos (x)} d x = \int \frac{sin (x)}{{cos}^{4} (x)} d x =$ $int1/cos^3(x)*sin(x)/cos(x)dx=intsin(x)/cos^4(x)dx=$

Now: $d [cos (x)] = - sin (x) d x$ $d[cos(x)]=-sin(x)dx$

I can write the integral in a new equivalent form:

$- \int \frac{d [cos (x)]}{{cos}^{4} (x)} =$ $-int(d[cos(x)])/cos^4(x)=$ now you can use $cos (x)$ $cos(x)$ as if it were a simple $x$ $x$ during your integration, giving:

$- \int \frac{d [cos (x)]}{{cos}^{4} (x)} = - \int {cos}^{- 4} (x) d [cos (x)] =$ $-int(d[cos(x)])/cos^4(x)=-intcos^(-4)(x)d[cos(x)]=$
(as for $- \int x^{- 4} d x$ $-intx^-4dx$ )

$= \frac{{cos}^{- 3}}{3} + c = \frac{1}{3 {cos}^{3} (x)} + c$ $=cos^(-3)/3+c=1/(3cos^3(x))+c$

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

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