Question

ALGUIEN SABE COMO SE HACE ESTA INTEGRAL DOY 40 PUNTOS:$\int _{\frac{\pi }{6\:}}^{\frac{\pi }{4}\:}\:\frac{sin\left(x\right)dx}{cos^3\left(x\right)-sin^3\left(x\right)}$

Answer 1

Hagamos la sustitución: $x=\arctan u$ entonces
 
              $dx = \dfrac{1}{1+u^2}\,du\\ \\ \\ \sin x =\dfrac{u}{\sqrt{1+u^2}}\\ \\ \\ \cos x = \dfrac{1}{\sqrt{1+u^2}}$

Sustituyamos

$\displaystyle I=\int_{\tan (\pi/6)}^{\tan(\pi/4)} \dfrac{u}{1-u^3}\,du\\ \\ \\ I=\int_{1/\sqrt{3}}^{1/\sqrt{2}} \dfrac{u}{3(u^2+u+1)}-\dfrac{1}{3(u^2+u+1)}-\dfrac{1}{3(u-1)}$

$\displaystyle I=\dfrac{1}{6}\int_{1/\sqrt{3}}^{1/\sqrt{2}}\dfrac{2u+1}{u^2+u+1}-\dfrac{3}{u^2+u+1}\, du-\dfrac{1}{3}\left.\left(\ln |u-1|\right)\right|_{1/\sqrt{3}}^{1/\sqrt{2}}\\ \\ \\ I=\dfrac{1}{6}\left.\left[\ln (u^2+u+1)\right]\right|_{1/\sqrt{3}}^{1/\sqrt{2}}-\dfrac{1}{2}\int_{1/\sqrt{3}}^{1/\sqrt{2}}\dfrac{du}{\left(u+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}-\cdots\\ \\ \cdots-\dfrac{1}{3}\ln \left|\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{2}}\right|$


$I=\dfrac{1}{6}\ln\left(\dfrac{9+3\sqrt{2}}{8+2\sqrt{3}}\right)-\dfrac{1}{\sqrt{3}}\left[\arctan\left(\dfrac{2u+1}{\sqrt{3}}\right)\right]_{1/\sqrt{3}}^{1/\sqrt{2}}-\dfrac{1}{3}\ln\left(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{2}}\right)$

De allí continua con el arcotangente...