ALGUIEN SABE COMO SE HACE ESTA INTEGRAL DOY 40 PUNTOS:![\int _{\frac{\pi }{6\:}}^{\frac{\pi }{4}\:}\:\frac{sin\left(x\right)dx}{cos^3\left(x\right)-sin^3\left(x\right)} \int _{\frac{\pi }{6\:}}^{\frac{\pi }{4}\:}\:\frac{sin\left(x\right)dx}{cos^3\left(x\right)-sin^3\left(x\right)}](https://tex.z-dn.net/?f=%5Cint+_%7B%5Cfrac%7B%5Cpi+%7D%7B6%5C%3A%7D%7D%5E%7B%5Cfrac%7B%5Cpi+%7D%7B4%7D%5C%3A%7D%5C%3A%5Cfrac%7Bsin%5Cleft%28x%5Cright%29dx%7D%7Bcos%5E3%5Cleft%28x%5Cright%29-sin%5E3%5Cleft%28x%5Cright%29%7D)
thomascaycedo19:
seguro de que está bien escrita?
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Hagamos la sustitución:
entonces
![dx = \dfrac{1}{1+u^2}\,du\\ \\ \\
\sin x =\dfrac{u}{\sqrt{1+u^2}}\\ \\ \\
\cos x = \dfrac{1}{\sqrt{1+u^2}} dx = \dfrac{1}{1+u^2}\,du\\ \\ \\
\sin x =\dfrac{u}{\sqrt{1+u^2}}\\ \\ \\
\cos x = \dfrac{1}{\sqrt{1+u^2}}](https://tex.z-dn.net/?f=dx+%3D+%5Cdfrac%7B1%7D%7B1%2Bu%5E2%7D%5C%2Cdu%5C%5C+%5C%5C+%5C%5C%0A%5Csin+x+%3D%5Cdfrac%7Bu%7D%7B%5Csqrt%7B1%2Bu%5E2%7D%7D%5C%5C+%5C%5C+%5C%5C%0A%5Ccos+x+%3D+%5Cdfrac%7B1%7D%7B%5Csqrt%7B1%2Bu%5E2%7D%7D)
Sustituyamos
![\displaystyle
I=\int_{\tan (\pi/6)}^{\tan(\pi/4)} \dfrac{u}{1-u^3}\,du\\ \\ \\
I=\int_{1/\sqrt{3}}^{1/\sqrt{2}} \dfrac{u}{3(u^2+u+1)}-\dfrac{1}{3(u^2+u+1)}-\dfrac{1}{3(u-1)}
\displaystyle
I=\int_{\tan (\pi/6)}^{\tan(\pi/4)} \dfrac{u}{1-u^3}\,du\\ \\ \\
I=\int_{1/\sqrt{3}}^{1/\sqrt{2}} \dfrac{u}{3(u^2+u+1)}-\dfrac{1}{3(u^2+u+1)}-\dfrac{1}{3(u-1)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%0AI%3D%5Cint_%7B%5Ctan+%28%5Cpi%2F6%29%7D%5E%7B%5Ctan%28%5Cpi%2F4%29%7D+%5Cdfrac%7Bu%7D%7B1-u%5E3%7D%5C%2Cdu%5C%5C+%5C%5C+%5C%5C%0AI%3D%5Cint_%7B1%2F%5Csqrt%7B3%7D%7D%5E%7B1%2F%5Csqrt%7B2%7D%7D+%5Cdfrac%7Bu%7D%7B3%28u%5E2%2Bu%2B1%29%7D-%5Cdfrac%7B1%7D%7B3%28u%5E2%2Bu%2B1%29%7D-%5Cdfrac%7B1%7D%7B3%28u-1%29%7D%0A)
![\displaystyle
I=\dfrac{1}{6}\int_{1/\sqrt{3}}^{1/\sqrt{2}}\dfrac{2u+1}{u^2+u+1}-\dfrac{3}{u^2+u+1}\, du-\dfrac{1}{3}\left.\left(\ln |u-1|\right)\right|_{1/\sqrt{3}}^{1/\sqrt{2}}\\ \\ \\
I=\dfrac{1}{6}\left.\left[\ln (u^2+u+1)\right]\right|_{1/\sqrt{3}}^{1/\sqrt{2}}-\dfrac{1}{2}\int_{1/\sqrt{3}}^{1/\sqrt{2}}\dfrac{du}{\left(u+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}-\cdots\\ \\
\cdots-\dfrac{1}{3}\ln \left|\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{2}}\right|\\ \\ \\
\displaystyle
I=\dfrac{1}{6}\int_{1/\sqrt{3}}^{1/\sqrt{2}}\dfrac{2u+1}{u^2+u+1}-\dfrac{3}{u^2+u+1}\, du-\dfrac{1}{3}\left.\left(\ln |u-1|\right)\right|_{1/\sqrt{3}}^{1/\sqrt{2}}\\ \\ \\
I=\dfrac{1}{6}\left.\left[\ln (u^2+u+1)\right]\right|_{1/\sqrt{3}}^{1/\sqrt{2}}-\dfrac{1}{2}\int_{1/\sqrt{3}}^{1/\sqrt{2}}\dfrac{du}{\left(u+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}-\cdots\\ \\
\cdots-\dfrac{1}{3}\ln \left|\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{2}}\right|\\ \\ \\](https://tex.z-dn.net/?f=%5Cdisplaystyle%0AI%3D%5Cdfrac%7B1%7D%7B6%7D%5Cint_%7B1%2F%5Csqrt%7B3%7D%7D%5E%7B1%2F%5Csqrt%7B2%7D%7D%5Cdfrac%7B2u%2B1%7D%7Bu%5E2%2Bu%2B1%7D-%5Cdfrac%7B3%7D%7Bu%5E2%2Bu%2B1%7D%5C%2C+du-%5Cdfrac%7B1%7D%7B3%7D%5Cleft.%5Cleft%28%5Cln+%7Cu-1%7C%5Cright%29%5Cright%7C_%7B1%2F%5Csqrt%7B3%7D%7D%5E%7B1%2F%5Csqrt%7B2%7D%7D%5C%5C+%5C%5C+%5C%5C%0AI%3D%5Cdfrac%7B1%7D%7B6%7D%5Cleft.%5Cleft%5B%5Cln+%28u%5E2%2Bu%2B1%29%5Cright%5D%5Cright%7C_%7B1%2F%5Csqrt%7B3%7D%7D%5E%7B1%2F%5Csqrt%7B2%7D%7D-%5Cdfrac%7B1%7D%7B2%7D%5Cint_%7B1%2F%5Csqrt%7B3%7D%7D%5E%7B1%2F%5Csqrt%7B2%7D%7D%5Cdfrac%7Bdu%7D%7B%5Cleft%28u%2B%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E2%2B%5Cdfrac%7B3%7D%7B4%7D%7D-%5Ccdots%5C%5C+%5C%5C%0A%5Ccdots-%5Cdfrac%7B1%7D%7B3%7D%5Cln+%5Cleft%7C%5Cdfrac%7B%5Csqrt%7B6%7D-%5Csqrt%7B3%7D%7D%7B%5Csqrt%7B6%7D-%5Csqrt%7B2%7D%7D%5Cright%7C%5C%5C+%5C%5C+%5C%5C%0A)
![I=\dfrac{1}{6}\ln\left(\dfrac{9+3\sqrt{2}}{8+2\sqrt{3}}\right)-\dfrac{1}{\sqrt{3}}\left[\arctan\left(\dfrac{2u+1}{\sqrt{3}}\right)\right]_{1/\sqrt{3}}^{1/\sqrt{2}}-\dfrac{1}{3}\ln\left(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{2}}\right)\\ \\ \\
I=\dfrac{1}{6}\ln\left(\dfrac{9+3\sqrt{2}}{8+2\sqrt{3}}\right)-\dfrac{1}{\sqrt{3}}\left[\arctan\left(\dfrac{2u+1}{\sqrt{3}}\right)\right]_{1/\sqrt{3}}^{1/\sqrt{2}}-\dfrac{1}{3}\ln\left(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{2}}\right)\\ \\ \\](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7B1%7D%7B6%7D%5Cln%5Cleft%28%5Cdfrac%7B9%2B3%5Csqrt%7B2%7D%7D%7B8%2B2%5Csqrt%7B3%7D%7D%5Cright%29-%5Cdfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D%5Cleft%5B%5Carctan%5Cleft%28%5Cdfrac%7B2u%2B1%7D%7B%5Csqrt%7B3%7D%7D%5Cright%29%5Cright%5D_%7B1%2F%5Csqrt%7B3%7D%7D%5E%7B1%2F%5Csqrt%7B2%7D%7D-%5Cdfrac%7B1%7D%7B3%7D%5Cln%5Cleft%28%5Cdfrac%7B%5Csqrt%7B6%7D-%5Csqrt%7B3%7D%7D%7B%5Csqrt%7B6%7D-%5Csqrt%7B2%7D%7D%5Cright%29%5C%5C+%5C%5C+%5C%5C%0A)
De allí continua con el arcotangente...
Sustituyamos
De allí continua con el arcotangente...
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