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I came across this problem: $$\int_0^\frac{\pi}{4} \frac{x \cos^2(2x)}{(1+\sin(2x)) (1+ \cos(2x))} dx$$

My attempt: $$\int_0^\frac{\pi}{4} \frac{x \cos^2(2x)}{(1+\sin(2x)) (1+ \cos(2x))} dx dx= \int_0^\frac{\pi}{4} \frac{x (1-\sin(2x))}{ 1+\cos(2x)} dx= \frac{1}{4}\int_0^\frac{\pi}{2} \frac{x (1-\sin(x))}{1+\cos(x) }dx$$

$$= \frac{1}{4} \left( \int_0^\frac{\pi}{2} \frac{x }{1+\cos(x) }dx- \int_0^\frac{\pi}{2} \frac{x \sin(x)}{1+\cos(x) }\right)dx $$

Here I got stuck, I couldn't evaluate those two integrals.

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  • $\begingroup$ I don't know if it will work out, but if you multiply the top and the bottom both by $1-\cos(x)$ and use the pytheagorean identity, you get something much simpler looking. The factor of x makes me suspect integration by parts will be helpful as well. $\endgroup$
    – Aaron
    Apr 6 at 20:52

4 Answers 4

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I'll just do the last two integrals you got to:

For the first you can apply IBP choosing $u=x$ and $\mathrm dv=\dfrac{\mathrm dx}{1+\cos x}$: $$\begin{aligned} I_1\equiv\int_0^{\pi/2}\!\!\!\!\frac{x}{1+\cos x}\,\mathrm dx&\overset{(\ast)^1}{=}\underbrace{\left[x\tan\left(\frac{x}{2}\right)\right|_0^{\pi/2}}_{\pi/2}-\int_0^{\pi/2}\!\!\!\!\tan\left(\frac{x}{2}\right)\,\mathrm dx\\ &\overset{(\ast)^2}{=}\frac{\pi}{2}-\left[-2\ln\left(\cos\left(\frac{x}{2}\right)\right)\right|_0^{\pi/2}=\frac{\pi}{2}-\ln 2 \end{aligned}$$ As for the other one you can apply IBP yet again by choosing $u=x$ and $\mathrm dv=\dfrac{\sin x\ \mathrm dx}{1+\cos x}$: $$\begin{aligned} I_2\equiv\int_0^{\pi/2}\!\!\!\!\frac{x\sin x}{1+\cos x}\,\mathrm dx&\overset{(\ast)^3}{=}\underbrace{\left[-x\ln(1+\cos x)\right|_0^{\pi/2}}_0+\int_0^{\pi/2}\!\!\!\!\ln(1+\cos x)\,\mathrm dx\\ &=2\mathrm G-\frac{\pi}{2}\ln2\\ \end{aligned}$$ where $\mathrm G$ is Catalan's constant and to compute the later integral you can adress to this MSE post.

Thus, $$ \int_0^{\pi/4}\!\!\!\!\frac{x\cos^2(2x)}{(1+\sin(2x))(1+\cos(2x))}\,\mathrm dx=\frac{1}{4}(I_2-I_1)=\boxed{\frac{\mathrm G}{2}+\frac{2-\pi}{8}\ln2-\frac{\pi}{8}} $$


$(\ast)^1:$ $$\int\! \dfrac{\mathrm dx}{1+\cos x}=\int\dfrac{1}{1+\dfrac{1-t^2}{1+t^2}}\frac{2\ \mathrm dt}{1+t^2}=\int\mathrm dt=t=\tan\left(\frac{x}{2}\right)$$ where here we used Weierstrass substitution $\left(t=\tan\left(\frac{x}{2}\right)\right)$ and so $\cos x=\frac{1-t^2}{1+t^2}$ and $\mathrm dx=\frac{2\mathrm dt}{1+t^2}$.

$(\ast)^2:$ $$\int \tan\left(\frac{x}{2}\right)\,\mathrm dx=\int\frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)}\,\mathrm dx=-2\int \frac{\mathrm du}{u}=-2\ln u=-2\ln\left(\cos\left(\frac{x}{2}\right)\right)$$ where we did $u=\cos\left(\frac{x}{2}\right)\iff -2\mathrm du=\sin\left(\frac{x}{2}\right)\,\mathrm dx$.

$(\ast)^3:$ $$\int\! \dfrac{\sin x\ \mathrm dx}{1+\cos x}=-\int\frac{\mathrm du}{u}=-\ln(1+\cos x)$$ where we did $u=1+\cos x\iff \mathrm du=-\sin x\,\mathrm dx$.

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Im giving a more rigorous trigonometrical answer.

$$f(x)=\frac{x\cos^2 (2x)}{(1+\sin(2x))(1+\cos (2x))}$$ $$=\frac{x\cos^2 (2x)}{(\sin x+\cos x)^2(2\cos^2 x)}$$ $$=\frac{x}{2}\bigg(\frac{\cos (2x)}{(\sin x+\cos x)(\cos x)}\bigg)^2$$ $$=\frac{x}{2}\bigg(\frac{(\cos x+ \sin x)(\cos x - \sin x)}{(\sin x+\cos x)(\cos x)}\bigg)^2$$ $$=\frac{x}{2}(1-\tan x)^2$$ Now $$\int_{0}^{\frac{π}{4}} \frac{x}{2}(1-\tan x)^2dx=\frac{1}{2}\int_{0}^{\frac{π}{4}} (x\sec^2x - 2x\tan x) \ dx$$ $$=\frac{1}{2}\int_{0}^{\frac{π}{4}} x \sec^2 x \ dx -\int_{0}^{\frac{π}{4}} x \tan x \ dx$$ Using integration by parts, this is $$=\frac{1}{2}\bigg(\frac{π}{4} - \frac{\ln 2}{2}\bigg) -\bigg(\frac{C'}{2} - \frac{π}{8} \ln 2\bigg)+C$$ Here $C'$=Catalan's constant.This post explains it well

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Note $$\int\frac{\cos^2 (2x)}{(\sin x+\cos x)^2(2\cos^2 x)}dx=\frac12\int(1-\tan x)^2dx=\ln(\cos x)+\frac12\tan(x)+C$$ and so $$\begin{eqnarray} &&\int_{0}^{\frac{π}{4}} \frac{x}{2}(1-\tan x)^2dx\\ &=&\int_{0}^{\frac{π}{4}}xd(\ln(\cos x)+\frac12\tan(x))\\ &=&x(\ln(\cos x)+\frac12\tan(x))\bigg|_0^{\frac\pi4}-\int_{0}^{\frac{π}{4}}(\ln(\cos x)+\frac12\tan(x)) dx\\ &=&\frac\pi8-\frac14\ln2+\frac1{8}\pi\ln2-\frac12C. \end{eqnarray}$$ Here $$ 2\int_{0}^{\frac{π}{4}}\ln(2\cos x)dx=C $$ is used from https://brilliant.org/wiki/catalans-constant/.

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$$ \begin{aligned} I & =\int_0^{\frac{\pi}{4}} \frac{x(1-\sin (2 x))}{1+\cos (2 x)} d x \\ & =\int_0^{\frac{\pi}{4}} \frac{x(1-2 \sin x \cos x)}{2 \cos ^2 x} d x \\ & =\frac{1}{2} \underbrace{\int_0^{\frac{\pi}{4}} \frac{x}{\cos ^2 x} d x}_{J} -\ \underbrace{ \int_0^{\frac{\pi}{4}} x \tan x d x}_{K} \end{aligned} $$ $$ \begin{aligned} J & =\frac{1}{2} \int_0^{\frac{\pi}{4}} x d(\tan x)\\ & =\frac{1}{2}\left([x \tan x]_0^{\frac{\pi}{4}}-\int_0^{\frac{\pi}{4}} \tan x d x\right)\\&= \frac{\pi}{4}-\frac{1}{2} \ln 2 \end{aligned} $$ $$ \begin{aligned} K & =-\int_0^{\frac{\pi}{4}} x d(\ln (\cos x)) \\ & =-[x \ln (\cos x)]_0^{\frac{\pi}{4}}+\int_0^{\frac{\pi}{4}} \ln (\cos x) d x \cdots (*)\\ & =-\frac{\pi}{4} \ln \left(\frac{1}{\sqrt{2}}\right)+\frac{G}{2}-\frac{\pi}{4} \ln 2 \\ & =\frac{\pi}{8} \ln 2+\frac{G}{2}-\frac{\pi}{4} \ln 2 \end{aligned} $$ where (*) uses the result.

Putting back yields $$ I=\frac{\pi}{8}-\frac{1}{4} \ln 2+\frac{\pi}{8} \ln 2-\frac{G}{2} $$ where $G$ is the Catalan’s constant.

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