How do you integrate $\int \frac{sin x}{{cos}^{3} x} d x$ $int sinx/cos^3xdx$ ?

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How do you integrate $\int \frac{sin x}{{cos}^{3} x} d x$ $int sinx/cos^3xdx$ ?

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Answer 1 · 2016-11-22T14:05:25

The derivative of $cos x$ $cosx$ is $- sin x$ $-sinx$ . Use substitution.

Explanation:

$\int \frac{sin x}{{cos}^{3} x} d x = - \int {(cos x)}^{- 3} (- sin x) d x$ $int sinx/cos^3x dx = - int (cosx)^-3 (-sinx) dx$

$= - \int u^{- 3} d u$ $= -int u^-3 du$

$= - \frac{u^{- 2}}{- 2} + C = \frac{1}{2 u^{2}} + C$ $= -u^-2/(-2) +C = 1/(2u^2) +C$

$= \frac{1}{2 {cos}^{2} x} + C$ $=1/(2cos^2 x) +C$

$= \frac{1}{2} {sec}^{2} x + C$ $=1/2sec^2x + C$

(Check the answer by differentiation.)

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