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Intergration question ~ Help

So I found this question, and I can't do it, even it seems quite simple.

Intergrate: 1/(cosx+sinx) dx

Showing me the steps how you did it would be awesome, please don't give me just an answer.

Thanks~ :smile:

Reply 1

user2020user

Original post by Cashmee

...

Are there any bounds for the integral?

(edited 9 years ago)

Reply 2

Cashmee

Original post by Khallil

Are there any bounds for the integral?

Nope, just integrate it, so you leave it with a +c on the end.

Reply 3

user2020user

Original post by Cashmee

Nope, just integrate it, so you leave it with a +c on the end.

I'd recommend a

t = \tan \frac{x}{2}

substitution if you (or I) can't find a more ingenious method.

Reply 4

Notnek

Original post by Cashmee

Write the denominator in the form

\displaystyle R\sin\left(x+\alpha\right )

then use a result that should be in your formula booklet.

Reply 5

Cashmee

Original post by notnek

Write the denominator in the form

\displaystyle R\sin\left(x+\alpha\right )

then use a result that should be in your formula booklet.

Hmm..ok I got.

sinx+cosx = Rsin(x+a)
sinx+cosx = Rsinxcosa + Rcosxsina

1= Rcosa
1= Rsina

Or 1=tana

What do I do now?

Reply 6

Cashmee

Original post by Khallil

I'd recommend a

t = \tan \frac{x}{2}

substitution if you (or I) can't find a more ingenious method.

Hmm..I don't think we been taught that method in C4.
Also I don't do further maths, so some integration techniques I do not know in FP1, FP2 and FP3.

Reply 7

user2020user

Original post by Cashmee

...

I see. You can maybe try it in your spare time to get a grips with the sub. It's called the Weierstrass substitution and is used to manipulate rational functions of sines and cosines into forms that come in handy for a whole host of things (including integrals).

For your working out in the post above, I'd suggest finding

R

in addition to

\alpha

Unparseable latex formula:

\begin{aligned} \left \begin{matrix} R\cos \alpha = 1 \\ R \sin \alpha = 1 \end{matrix} \ \right\} \implies R^2 = 2 \ \ \& \ \ \alpha = \arctan (1) \end{aligned}

Remember that the arctangent function yields the principle value of

\alpha

which is

-\pi < \alpha \leq \pi

(edited 9 years ago)

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Intergration question ~ Help

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