Question

Calcule a integral indefinida abaixo:

Answer 1

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$\displaystyle\sf\int cos^6(3x)~dx=\int(cos^2(3x))^2\cdot cos^2(3x)~dx\\\displaystyle\sf\int \left(\dfrac{1}{2}[1+cos(6x)]\right)^2\cdot\dfrac{1}{2}[1+cos(6x)]~dx\\\displaystyle\sf\int\dfrac{1}{4}[1+2cos(6x)+cos^2(6x)]\cdot\dfrac{1}{2}[1+cos(6x)]~dx\\\displaystyle\sf\dfrac{1}{8}\int[1+cos(6x)+2cos(6x)+2cos^2(6x)+cos^2(6x)+cos^3(6x)]~dx$

$\displaystyle\sf\dfrac{1}{8}\int[1+3cos(6x)+3cos^2(6x)+cos^3(6x)]\\\displaystyle\sf\dfrac{1}{8}\int dx+\dfrac{3}{8}\int cos(6x)~dx+\dfrac{3}{8}\int cos^2(6x)~dx+\dfrac{1}{8}\int cos^3(6x)~dx$

$\displaystyle\sf\dfrac{1}{8}\int dx=\dfrac{1}{8}x+c_1\\\displaystyle\sf\dfrac{3}{8}\int cos(6x)~dx=\dfrac{\diagup\!\!\!3}{8\cdot\diagup\!\!\!6_2}sen(6x)+c_2=\dfrac{1}{16}sen(6x)+c_2$

$\displaystyle\sf\dfrac{3}{8}\int cos^2(6x)~dx=\dfrac{3}{8}\cdot\dfrac{1}{2}\int[1+cos(12x)]~dx\\\displaystyle\sf\dfrac{3}{16}\int dx+\dfrac{3}{16}\int cos(12x)~dx=\dfrac{3}{16}x+\dfrac{\diagup\!\!\!3}{16\cdot\diagup\!\!\!\!12_4}sen(12x)+c_3\\\displaystyle\sf\dfrac{3}{8}\int cos^2(6x)~dx=\dfrac{3}{16}x+\dfrac{1}{48}sen(12x)+c_3$

$\displaystyle\sf\dfrac{3}{8}\int cos^3(6x)~dx=\dfrac{3}{8}\int cos^2(6x)\cdot cos(6x)~dx\\\displaystyle\sf\dfrac{3}{8}\int(1-sen^2(6x))\cdot cos(6x)~dx=\dfrac{3}{8}\cdot\dfrac{1}{6}\int(1-sen^2(6x))\cdot cos(6x)\cdot6~dx\\\sf fac_{\!\!,}a~u=sen(6x)\implies du= cos(6x)\cdot 6~dx\\\displaystyle\sf\dfrac{\diagdown\!\!\!3}{8}\cdot\dfrac{1}{\diagdown\!\!\!6_2}\int(1-sen^2(6x))\cdot cos(6x)\cdot 6~dx=\dfrac{1}{16}\int(1-u^2)~du\\\sf \dfrac{1}{16}u-\dfrac{1}{48}u^3+c_4$

$\displaystyle\sf\dfrac{3}{8}\int cos^3(6x)~dx=\dfrac{1}{16}sen(6x)-\dfrac{1}{48}sen^3(6x)+c_4$

$\displaystyle\sf\int cos^6(3x)~dx\\\sf =\dfrac{1}{8}x+\dfrac{1}{16}sen(6x)+\dfrac{3}{16}x+\dfrac{1}{48}sen(12x)+\dfrac{1}{16}sen(6x)-\dfrac{1}{48}sen^3(6x)+k\\\displaystyle\sf\int cos^6(3x)~dx=\dfrac{5}{16}x+\dfrac{1}{8}sen(6x)+\dfrac{1}{48}sen(12x)-\dfrac{1}{48}sen^3(6x)+k$