- #1

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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

- Thread starter velliyangiri
- Start date

- #1

- 1

- 0

∫ 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

- #2

- 658

- 2

- #3

- #4

- 658

- 2

Lol.. Oh my..

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

- #5

- 2,967

- 5

[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

- 349

- 1

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

[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]

- #7

- 658

- 2

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.

[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]

- #8

- 2,967

- 5

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

vela

Staff Emeritus

Science Advisor

Homework Helper

Education Advisor

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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

- 658

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

- #11

- 15,393

- 685

There is.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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