Question

ayuda con la siguiente integral

integral de sec^6xcos2xdx

Answer 1

$\displaystyle I=\int \sec^6\alpha\cos 2\alpha~d\alpha\\ \\ I=\int \sec^6\alpha(2\cos ^2 \alpha-1)~d\alpha\\ \\ I=\int 2\sec^4\alpha-\sec^6\alpha~d\alpha\\ \\ u=\sec \alpha\to du = \sec \alpha\tan \alpha~d\alpha=u\sqrt{u^2-1}~d\alpha\\ \\ d\alpha=\dfrac{du}{u\sqrt{u^2-1}}\\ \\ \\ I=\int (2u^4-u^6)\cdot \dfrac{du}{u\sqrt{u^2-1}}$

Tenemos esta integral al hacer sustituciones

$\displaystyle I=\int \dfrac{2u^3-u^5}{\sqrt{u^2-1}}du\\ \\ \\ \texttt{otra sustituci\'on}: t=\sqrt{u^2-1}\to t^2=u^2-1\to t~dt=u~du\\ \\ \\ I=\int \dfrac{2u^2-u^4}{\sqrt{u^2-1}}~u~du\\ \\ \\ I=\int\dfrac{2(t^2+1)-(t^2+1)^2}{t}\cdot t~dt\\ \\ \\ I=\int 2(t^2+1)-(t^2+1)^2~dt\\ \\ \\ I=\int 1-t^4~dt\\ \\ \\ I=t-\dfrac{t^5}{5}=\sqrt{u^2-1}-\dfrac{\sqrt{u^2-1}^5}{5}=\tan\alpha-\dfrac{\tan^5\alpha}{5}\\ \\ \\ \int \sec^6\alpha\cos 2\alpha~d\alpha=\tan\alpha-\dfrac{\tan^5\alpha}{5}+C$