What is $\int \frac{sin x}{{cos}^{2} x} d x$ $int(sinx)/(cos^2 x) dx$ ?

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What is $\int \frac{sin x}{{cos}^{2} x} d x$ $int(sinx)/(cos^2 x) dx$ ?

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Answer 1 · 2018-04-21T06:32:03

$sec x + C$ $secx+C$

Explanation:

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

using u-substitution:
let $u = cos x$ $u=cosx$ . then $d u = - sin x d x$ $du=-sinxdx$ or $- d u = sin x d x$ $-du=sinxdx$

substituting into the original integral:
$\int - \frac{1}{u^{2}} d u$ $int-1/u^2du$

integrate that with power rule:

$\frac{1}{u} + C$ $1/u+C$

substitute $u = cos x$ $u=cosx$ :

$\frac{1}{cos x} + C = sec x + C$ $1/cosx+C=secx+C$

Answer 2 · 2018-04-21T12:59:00

$sec x + C$ $secx+C$

Explanation:

Based on the "inverse" nature of derivatives and antiderivatives (found through integrals), if we know that $\frac{d}{d x} sec x = sec x tan x$ $d/dxsecx=secxtanx$ , then $\int sec x tan x d x = sec x + C$ $intsecxtanxdx=secx+C$ .

Here, using $\frac{1}{cos x} = sec x$ $1/cosx=secx$ and $tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$ , we see that

$\int \frac{sin x}{{cos}^{2} x} d x = \int \frac{1}{cos x} (\frac{sin x}{cos x}) d x = \int sec x tan x d x = sec x + C$ $intsinx/cos^2xdx=int1/cosx(sinx/cosx)dx=intsecxtanxdx=secx+C$

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What is $\int \frac{sin x}{{cos}^{2} x} d x$ $int(sinx)/(cos^2 x) dx$ ?

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Explanation:

Explanation:

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