How do you find the integral of $f (x) = \frac{1}{1 - sin x}$ $f(x)=1/(1-sinx)$ using integration by parts?

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How do you find the integral of $f (x) = \frac{1}{1 - sin x}$ $f(x)=1/(1-sinx)$ using integration by parts?

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Answer 1 · 2018-04-10T22:49:05

$tan x + sec x + C$ $tanx+secx+C$

Explanation:

$\int \frac{d x}{1 - sin x}$ $intdx/(1-sinx)$

Let's first get this into a more workable form by multiplying through by the conjugate:

$= \int \frac{1}{1 - sin x} \cdot \frac{1 + sin x}{1 + sin x} d x$ $=int1/(1-sinx)*(1+sinx)/(1+sinx)dx$

$= \int \frac{1 + sin x}{1 - {sin}^{2} x} d x$ $=int(1+sinx)/(1-sin^2x)dx$

Recall that ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$ :

$= \int \frac{1 + sin x}{{cos}^{2} x} d x$ $=int(1+sinx)/cos^2xdx$

Splitting up the integral by addition:

$= \int \frac{1}{{cos}^{2} x} d x + \int \frac{sin x}{{cos}^{2} x} d x$ $=int1/cos^2xdx+intsinx/cos^2xdx$

Rewrite using the identities $sec x = \frac{1}{cos x}$ $secx=1/cosx$ and $tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$ :

$= \int {sec}^{2} x d x + \int sec x tan x d x$ $=intsec^2xdx+intsecxtanxdx$

And then, from the knowledge that $\frac{d}{d x} tan x = {sec}^{2} x$ $d/dxtanx=sec^2x$ and $\frac{d}{d x} sec x = sec x tan x$ $d/dxsecx=secxtanx$ , we see that:

$= tan x + sec x + C$ $=tanx+secx+C$

No integration by parts required!

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How do you find the integral of $f (x) = \frac{1}{1 - sin x}$ $f(x)=1/(1-sinx)$ using integration by parts?

1 Answer

Explanation:

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