Question

Calcule a integral trigonometrica

Answer 1

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$\displaystyle\sf\int tg^6x~sec^4x~dx=\int tg^6x~sec^2x~sec^2x~dx\\\displaystyle\sf\int tg^6x(1+tg^2x)~sec^2x~dx\\\underline{\rm fac_{\!\!,}a}\\\sf u=tgx\implies du=sec^2x~dx\\\displaystyle\sf\int tg^6x(1+tg^2x)~sec^2x~dx=\int u^6(1+u^2)du\\\displaystyle\sf\int(u^6+u^8)~du=\dfrac{1}{7}u^7+\dfrac{1}{9}u^9+k\\\boxed{\displaystyle\sf\int tg^6x~sec^4x~dx=\dfrac{1}{7}tg^7x+\dfrac{1}{9}tg^9x+k}$