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$$\int_0^1 \frac 1 {1+y\cos(x)}dx$$

If there was no $y$, I would multiply by $1-\cos(x)$ and finish it quickly. But the $y$ stops me from doing that. I tried the trigonometric substitution but failed. I assume there is a simple way to solve this. I would appreciate if anyone could help me with this. Thanks.

EDIT: When trying the Weirstrass Sub. I got here and wasn't able to find a way to go forward:

$$\int_0^{\pi/4}\frac {2dt}{1+y+t^2(1-y)}$$

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  • $\begingroup$ Did you consider to myltiply and divide by $1-y\cos x$? If so, what is your problem with the result? (Also, the upper limit is a bit strange for cosine, did this integral come from an exercise or from some application?) $\endgroup$ – mickep Feb 7 '17 at 15:23
  • $\begingroup$ You can see $y$ as a constant, so you can multiply and divide by $1-ycos(x)$ and continue with your way. In fact the only variable is $x$. $\endgroup$ – Giulio Feb 7 '17 at 15:23
  • $\begingroup$ @Giulio. That is not going to work here. The Pythagorean Theorem doesn't work anymore if $y$ is not $1$ or $-1$. $\endgroup$ – imranfat Feb 7 '17 at 15:25
  • $\begingroup$ @Ron. Have you looked into the Weierstrass substitution? $\endgroup$ – imranfat Feb 7 '17 at 15:25
  • $\begingroup$ @imranfat just noticed my typo $\endgroup$ – Giulio Feb 7 '17 at 15:28
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Hint:

  1. Substituting $t = \tan \frac{x}{2}$ gives $$I(y):=\int \frac{2}{1+y\cos x} \; \mathrm d x= \int \frac{2}{1+y \frac{1-t^2}{1+t^2}} \cdot \frac{2 \; \mathrm d t}{1+t^2} = \int \frac{2}{1+y+t^2(1-y)} \; \mathrm d t .$$

  2. Since $$\int \frac{1}{1+t^2} \; \mathrm d t = \arctan t$$ the final substitution should be easy.

  3. The last integral is of the type $$ \int \frac{1}{a +bt^2} \; \mathrm d t $$ with constants $a,b \in \mathbf R$. Now substitute $\frac{\sqrt{b}}{\sqrt{a}}t= z$ and you will get $$ \int \frac{1}{a +bt^2} \; \mathrm d t = \frac{1}{\sqrt{a}\sqrt{b}} \; \arctan \left( \frac{\sqrt{b}}{\sqrt{a}}t \right) .$$

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  • $\begingroup$ typo in your substitution. Should be $t=\tan\frac{x}{2}$? $\endgroup$ – Juniven Feb 7 '17 at 15:38
  • $\begingroup$ Yes, sure. Thank you! $\endgroup$ – Niklas Feb 7 '17 at 15:39
  • $\begingroup$ Thank you. I'm afraid my problem is exactly with that easy substitution, since I reached the same integral using the substitution. $\endgroup$ – Ron Feb 7 '17 at 15:44
  • $\begingroup$ @Ron, okay. Next hint added! $\endgroup$ – Niklas Feb 7 '17 at 15:54
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First one should be careful to consider only $y$ such that $1+y\cos x\neq 0$.

If that is the case we notice that $$ \frac{1}{1+y\cos x}=\frac{1-y\cos x}{1-y^2\cos^2x}. $$ Next, we divide into the parts $$ \frac{1}{1-y^2\cos^2x}=\frac{1}{\sin^2 x+(1-y^2)\cos^2x}=\frac{1}{\cos^2x}\frac{1}{\tan^2x+1-y^2} $$ For this part, let $u=\tan x$. For the second part, $$ \frac{-y\cos x}{1-y^2\cos^2x}=-\frac{y \cos x}{1-y^2+y^2\sin^2x}. $$ Here, let $u=\sin x$.

I think you are now on safe ground. Ask if you cannot take it from here.

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}{\dd x \over 1 + y\cos\pars{x}} & = \int_{0}^{1}{\dd x \over 1 + y\bracks{2\cos^{2}\pars{x/2} - 1}} = 2\int_{0}^{1/2}{\dd x \over 1 - y + 2y\cos^{2}\pars{x}} \\[5mm] & = 2\int_{0}^{1/2}{\sec^{2}\pars{x} \over \pars{1 - y}\sec^{2}\pars{x} + 2y}\,\dd x = 2\int_{0}^{1/2} {\sec^{2}\pars{x} \over \pars{1 - y}\tan^{2}\pars{x} + 1 + y}\,\dd x \\[5mm] & = 2\,{1 \over 1 + y}\,\root{1 + y \over 1 - y}\int_{0}^{1/2} {\root{\pars{1 - y}/\pars{1 + y}}\sec^{2}\pars{x} \over \bracks{\root{\pars{1 - y}/\pars{1 + y}}\tan\pars{x}}^{2} + 1}\,\dd x \\[5mm] & = {2 \over \root{1 - y^{2}}}\, \arctan\pars{\root{1 - y \over 1 + y}\tan\pars{1 \over 2}} \end{align}

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