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## Homework Statement

**∫ 1 /( sin x + sec x) dx**

## Homework Equations

## The Attempt at a Solution

∫ cos x / ( sin x + cos x ) style question

Tried ∫ cos x / (sin x . cos x + 1) dx

and uses sin 2x

tried substitutions

nothing seem to work

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So in summary, try substituting:\begin{align*}\frac{1}{\sin x + \sec x} &= \frac{1}{\sin x + \sec x} \cdot \frac{\sin x - \sec x}{\sin x - \sec x} \\&= \frac{\sin x - \sec x}{\sin^2 x - \sec^2 x} \\&= \frac{\cos^2 x\sin x - \cos x}{\cos^2 x\ \sin^2 x - 1} \\&= \frac{\cos^2 x}{\cos^2 x\ (1-\cos^2

- #1

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∫ cos x / ( sin x + cos x ) style question

Tried ∫ cos x / (sin x . cos x + 1) dx

and uses sin 2x

tried substitutions

nothing seem to work

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- #2

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- #4

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Highway said:

Lol.. Oh my..

There has to be an easier way to get this one.

- #5

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[tex]

t \equiv \tan \left( \frac{x}{2} \right)

[/tex]

[tex]

\sin x = \frac{2 t}{1 + t^2}, \ \cos x = \frac{1 - t^2}{1 + t^2}, \ dx = \frac{2 \, dt}{1 + t^2}

[/tex]

- #6

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Dickfore said:

[tex]

t \equiv \tan \left( \frac{x}{2} \right)

[/tex]

[tex]

\sin x = \frac{2 t}{1 + t^2}, \ \cos x = \frac{1 - t^2}{1 + t^2}, \ dx = \frac{2 \, dt}{1 + t^2}

[/tex]

man, i wish i knew how to do this back in the day. . .

- #7

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Dickfore said:

[tex]

t \equiv \tan \left( \frac{x}{2} \right)

[/tex]

[tex]

\sin x = \frac{2 t}{1 + t^2}, \ \cos x = \frac{1 - t^2}{1 + t^2}, \ dx = \frac{2 \, dt}{1 + t^2}

[/tex]

I can't see that being much easier because after playing with the algebra it's going to leave many t's with various higher degrees (up to 6th degree if my algebra is correct) and also in a fraction form. After plugging back in tan it's going to have some subs and then reduction formulas for those higher degrees.

- #8

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No one said the result should have a simple form. Given the integrand, no other simpler substitution seems possible. But, you need to do the partial fraction decomposition yourself, or at least bring the integral in a rational algebraic form if you want further help.

- #9

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\begin{align*}

\frac{1}{\sin x + \sec x} &= \frac{1}{\sin x + \sec x} \cdot \frac{\sin x - \sec x}{\sin x - \sec x} \\

&= \frac{\sin x - \sec x}{\sin^2 x - \sec^2 x} \\

&= \frac{\cos^2 x\sin x - \cos x}{\cos^2 x\ \sin^2 x - 1} \\

&= \frac{\cos^2 x}{\cos^2 x\ (1-\cos^2 x) - 1}\sin x - \frac{1}{(1-\sin^2 x)\sin^2 x - 1}\cos x

\end{align*}Then use a trig substitution for each of the two terms.

- #10

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Dickfore said:No one said the result should have a simple form. Given the integrand, no other simpler substitution seems possible. But, you need to do the partial fraction decomposition yourself, or at least bring the integral in a rational algebraic form if you want further help.

Yeah I know what you mean, it's just that the integral just looked deceiving simple at first glance.

- #11

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There is.DrummingAtom said:Lol.. Oh my..

There has to be an easier way to get this one.

Use the fact that cos(-pi/4) = sin(-pi/4)=1/√2 .

[tex]\begin{aligned}

\sin x + \cos x &= \sqrt2 \left(\cos x\frac 1 {\sqrt2} + \sin x \frac 1 {\sqrt2}\right) \\

&= \sqrt2 (\cos x \cos(\pi/4) + \sin x \sin (\pi/4)) \\

&= \sqrt2 \cos(x-\pi/4)

\end{aligned}[/tex]

And thus [itex]1/(\sin x + \cos x) = \sec(x-\pi/4)/\sqrt 2[/itex]

So now all you have to do is find the integral of sec(x-pi/4) and then divide by √2.

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