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I've to find this integral for a project I'm working on and I've no idea how to proceed, can someone help me out?

$$ \begin{align} & \int\left(\frac{1}{2}\left((\cos(2x)+1)\left(\frac{2\tanh^{-1}(x^2)} {\cos(2x)+1}+\cos(x)+\sin^2(x)\right)\right)\right)dx \\[10pt] = {} & \frac{1}{2}\int\left((\cos( 2x)+1)\left(\frac{2\tanh^{-1}(x^2)} {\cos(2x)+1}+\cos(x)+\sin^2(x)\right)\right)dx=\cdots \end{align} $$

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$$\int\left(\frac{1}{2}\left((\cos(2x)+1)\left(\frac{2\tanh^{-1}(x^2)}{\cos(2x)+1}+\cos(x)+\sin^2(x)\right)\right)\right)dx=$$ $$\frac{1}{2}\int\left((\cos(2x)+1)\left(\frac{2\tanh^{-1}(x^2)}{\cos(2x)+1}+\cos(x)+\sin^2(x)\right)\right)dx=$$ $$\frac{1}{2}\int\left((\cos(2x)+1)\left(\tanh^{-1}(x^2)\sec^2(x)+\cos(x)+\sin^2(x)\right)\right)dx=$$ $$\frac{1}{2}\int\left(2\cos^2(x)\left(\tanh^{-1}(x^2)\sec^2(x)+\cos(x)+\sin^2(x)\right)\right)dx=$$ $$\frac{1}{2}\cdot 2\int\left(\cos^2(x)\left(\tanh^{-1}(x^2)\sec^2(x)+\cos(x)+\sin^2(x)\right)\right)dx=$$ $$\int\left(\cos^2(x)\left(\tanh^{-1}(x^2)\sec^2(x)+\cos(x)+\sin^2(x)\right)\right)dx=$$ $$\int\left(\tanh^{-1}(x^2)+\cos^3(x)+\frac{1}{4}\sin^2(x)\right)dx=$$ $$\int\left(\tanh^{-1}(x^2)\right)dx+\int\left(\cos^3(x)\right)dx+\int\left(\frac{1}{4}\sin^2(x)\right)dx=$$ $$\int\tanh^{-1}(x^2)dx+\int\cos^3(x)dx+\frac{1}{4}\int\sin^2(x)dx=$$ $$\int\tanh^{-1}(x^2)dx+\int\cos^3(x)dx+\frac{1}{4}\int\left(\frac{1}{2}-\frac{1}{2}\cos(2x)\right)dx=$$ $$\int\tanh^{-1}(x^2)dx+\int\cos^3(x)dx+\frac{1}{4}\left(\frac{1}{2}\int\left(1\right)dx-\frac{1}{2}\int\left(\cos(2x)\right)dx\right)=$$ $$\int\tanh^{-1}(x^2)dx+\int\cos^3(x)dx+\frac{1}{4}\left(\frac{x}{2}-\frac{1}{2}\int\left(\cos(2x)\right)dx\right)=$$


Substitute $u=2x$ and $du=2dx$:


$$\int\tanh^{-1}(x^2)dx+\int\cos^3(x)dx+\frac{1}{4}\left(\frac{x}{2}-\frac{1}{4}\int\left(\cos(u)\right)du\right)=$$ $$\int\tanh^{-1}(x^2)dx+\int\cos^3(x)dx+\frac{1}{4}\left(\frac{x}{2}-\frac{\sin(u)}{4}\right)=$$ $$\int\tanh^{-1}(x^2)dx+\int\cos^3(x)dx+\frac{1}{4}\left(\frac{x}{2}-\frac{\sin(2x)}{4}\right)=$$ $$\int\tanh^{-1}(x^2)dx+\int\cos^3(x)dx+\frac{2x-\sin(2x)}{16}=$$ $$\int\tanh^{-1}(x^2)dx+\frac{\sin(x)\cos^2(x)}{3}+\frac{2}{3}\int\cos(x)dx+\frac{2x-\sin(2x)}{16}=$$ $$\int\tanh^{-1}(x^2)dx+\frac{\sin(x)\cos^2(x)}{3}+\frac{2\sin(x)}{3}+\frac{2x-\sin(2x)}{16}=$$ $$\int\tanh^{-1}(x^2)dx+\frac{9\sin(x)+\sin(3x)}{12}+\frac{2x-\sin(2x)}{16}=$$ $$x\tanh^{-1}(x^2)-\int\frac{2x^2}{1-x^4}dx+\frac{9\sin(x)+\sin(3x)}{12}+\frac{2x-\sin(2x)}{16}=$$ $$x\tanh^{-1}(x^2)-2\int\frac{x^2}{1-x^4}dx+\frac{9\sin(x)+\sin(3x)}{12}+\frac{2x-\sin(2x)}{16}=$$ $$x\tanh^{-1}(x^2)+2\int\frac{x^2}{x^4-1}dx+\frac{9\sin(x)+\sin(3x)}{12}+\frac{2x-\sin(2x)}{16}=$$ $$x\tanh^{-1}(x^2)+2\left(\int\left(\frac{1}{2(x^2+1)}-\frac{1}{4(x+1)}+\frac{1}{4(x-1)}\right)dx\right)+\frac{9\sin(x)+\sin(3x)}{12}+\frac{2x-\sin(2x)}{16}=$$ $$x\tanh^{-1}(x^2)+\frac{\ln|1-x|-\ln|x+1|+2\tan^{-1}(x)}{2}+\frac{9\sin(x)+\sin(3x)}{12}+\frac{2x-\sin(2x)}{16}+C$$

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  • 3
    $\begingroup$ That is impressive. When I first saw the integral I thought it would be intractable. $\endgroup$ – Dargscisyhp Aug 9 '15 at 18:17
  • 1
    $\begingroup$ this is a minor point, but I think in line 7 you want $\frac{1}{4}\sin^2{2x}$ $\endgroup$ – user84413 Aug 9 '15 at 20:16

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