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\[ I_1= \displaystyle\int\tan(x)\sqrt{2+\sqrt{4+\cos(x)}} dx \]
Primer sembla prudent d'eliminar les funcions trigonomètriques fent el canvi:
\[t= \cos x \hspace{24pt} dt=-\sin x dx\]
així
\[
I_1 = \displaystyle\int\dfrac{\sin(x)}{\cos(x)}\sqrt{2+\sqrt{4+\cos(x)}} dx
= \displaystyle\int\dfrac{\sqrt{2+\sqrt{4+\cos(x)}}}{\cos(x)} \sin(x)dx
= -\displaystyle\int\dfrac{\sqrt{2+\sqrt{4+t}}}{t} dt
\]
Ara fem el canvi
\[s=\sqrt{2+\sqrt{4+t}} \hspace{24pt} s^2=2+\sqrt{4+t} \hspace{24pt} (s^2-2)^2=4+t\]
llavors
\[t=(s^2-2)^2-4=s^4-4s^2+4-4=s^4-4s^2\]
si
\[t=s^4-4s^2=s^2(s^2-4) \hspace{24pt}\rightarrow\hspace{24pt} dt=(4s^3-8s)ds=4s(s^2-2)ds \]
i la integral
\[
I_1 = -\displaystyle\int\dfrac{s}{s^2(s^2-4)}4s(s^2-2)ds
\]
les variables \(s\) es cancelen i queda una integral molt més accessible
\[
I_1 = -4\displaystyle\int\dfrac{s^2-2}{s^2-4}ds
= -4\displaystyle\int\dfrac{s^2-4+2}{s^2-4}ds
= -4\displaystyle\int\left(1+\dfrac{2}{s^2-4}\right)ds
= -4s-8\displaystyle\int\dfrac{1}{s^2-4}ds
\]
és fàcil trobar que
\[\dfrac{1}{s^2-4}=\dfrac{1}{4}\left(\dfrac{1}{s-2}-\dfrac{1}{s+2}\right)\]
així
\[
I_1 = -4s-8\displaystyle\int\dfrac{1}{s^2-4}ds
= -4s-8\displaystyle\int\dfrac{1}{4}\left(\dfrac{1}{s-2}-\dfrac{1}{s+2}\right)ds
= -4s-2\displaystyle\int\dfrac{1}{s-2}ds+2\displaystyle\int\dfrac{1}{s+2}ds
\]
per tant la solució seria
\[
I_1
= -4s-2\log(s-2)+2\log(s+2)
= -4s+2\log\left(\dfrac{s+2}{s-2}\right)
\]
com que
\[s=\sqrt{2+\sqrt{4+t}} \hspace{24pt} t= \cos x \hspace{24pt} s=\sqrt{2+\sqrt{4+\cos x}}\]
la solució final
\[
I_1 = \displaystyle\int\tan(x)\sqrt{2+\sqrt{4+\cos(x)}} dx
= -4\sqrt{2+\sqrt{4+\cos x}}+2\log\left(\dfrac{\sqrt{2+\sqrt{4+\cos x}}+2}{\sqrt{2+\sqrt{4+\cos x}}-2}\right)
\]
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