Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am having some difficulties with the calculation of the following integral. Can somebody help me please?

$$\int \frac{dx}{1+a\cos x},\text{ for }0<a<1$$

Thank you in advance

share|cite|improve this question

closed as off-topic by RecklessReckoner, Stefan, zz20s, kamil09875, Steven Gregory Mar 20 at 17:58

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – RecklessReckoner, Stefan, zz20s, kamil09875, Steven Gregory
If this question can be reworded to fit the rules in the help center, please edit the question.

    
How many times are you going to ask the same question? – Mike Nov 7 '12 at 13:51
    
It is not the same question. It is another exercise.. – user43418 Nov 7 '12 at 13:52
    
It's the same basic integral rewritten. You've now had 3 people give you the same answer on 3 separate questions. – Mike Nov 7 '12 at 13:57

Substitute, $t = \tan \left(\dfrac x2\right)$. So $x = 2\tan^{-1}t$ and $dx = \dfrac{2dt}{1+t^2}$. And the integral becomes,

$$\begin{align*}\int\dfrac{\dfrac{2}{1+t^2}}{1+\dfrac{a(1-t^2)}{1+t^2}}dt &=\dfrac2{(1+a)}\int\dfrac1{1+\left[t\dfrac{\sqrt{1-a}}{\sqrt{1+a}}\right]^2}dt &\color{blue}{u =\left[t\frac{\sqrt{1-a}}{\sqrt{1+a}}\right]\Rightarrow dt = \frac{\sqrt{1+a}}{\sqrt{1-a}}du }\\ &=\dfrac2{(1+a)}\frac{\sqrt{1+a}}{\sqrt{1-a}}\int\dfrac1{1+u^2}du &\color{blue}{\int\dfrac1{1+u^2}du= \tan^{-1}u}\\ \\&=\dfrac2{(1+a)}\frac{\sqrt{1+a}}{\sqrt{1-a}}\tan^{-1}u\\\\ &=\dfrac2{\sqrt{1-a^2}}\tan^{-1}u&\color{blue}{u =\left[t\frac{\sqrt{1-a}}{\sqrt{1+a}}\right]}\\ \\ &=\dfrac{2\tan^{-1}\left[\frac{\sqrt{1-a}}{\sqrt{1+a}}\cdot\tan \left(\dfrac x2\right)\right]}{\sqrt{1-a^2}}\\\\ \end{align*}$$

$$\displaystyle\Large\therefore \boxed{\int \dfrac1{1+acosx}dx =\dfrac{2\tan^{-1}\left[\frac{\sqrt{1-a}}{\sqrt{1+a}}\cdot\tan \left(\dfrac x2\right)\right]}{\sqrt{1-a^2}} }$$

share|cite|improve this answer

Detailed hint:

Wikipedia calls this the The Weierstrass Substitutiion: when $t=\tan(\theta/2)$, $$ \begin{align} \sin(\theta)&=\frac{2t}{1+t^2}\\ \cos(\theta)&=\frac{1-t^2}{1+t^2}\\ \tan(\theta)&=\frac{2t}{1-t^2}\\ \mathrm{d}\theta&=\frac{2\mathrm{d}t}{1+t^2} \end{align} $$

share|cite|improve this answer

Hint 1: $\cos x=\frac{1-t^2}{1+t^2}$ which $t=\tan\frac{x}{2}$.

Hint 2: $t=\tan\frac{x}{2}\Rightarrow dt=\frac{1}{2}\sec^2\frac{x}{2}dx=\frac{1}{2}(t^2+1)dx$.

share|cite|improve this answer
1  
How do you replace the dx in integral expression ? I have a similar integral to calculate – Carpediem Nov 7 '12 at 13:55
    
So we have: $\int \frac{\frac{2}{t^2+1}}{1+a\frac{1-t^2}{1+t^2}}$ – user43418 Nov 7 '12 at 14:09

Not the answer you're looking for? Browse other questions tagged or ask your own question.