Question

integral sen^3 (1 − 2x). cos^3 (1 − 2x)dx

Answer 1

Calcular a integral indefinida:

     $\displaystyle\int \mathrm{sen^3}(1-2x)\cdot \cos^3(1-2x)\,dx$


Reescreva o integrando de forma conveniente:

     $\displaystyle=\int \frac{1}{-2}\cdot (-2)\,\mathrm{sen^3}(1-2x)\cdot \cos^3(1-2x)\,dx\\\\\\ =-\,\frac{1}{2}\int\mathrm{sen^3}(1-2x)\cdot \cos^3(1-2x)\,\cdot (-2)\,dx$


Como tanto o expoente do seno como do cosseno são ímpares, podemos escolher qual dos dois vamos decompor. Por exemplo, vamos decompor o cosseno:

     $\displaystyle=-\,\frac{1}{2}\int\mathrm{sen^3}(1-2x)\cdot \big[\cos^2(1-2x)\cdot \cos(1-2x)\big]\cdot (-2)\,dx\\\\\\ =-\,\frac{1}{2}\int\mathrm{sen^3}(1-2x)\cdot \cos^2(1-2x)\cdot \big[\!-2\cos(1-2x)\big]\,dx$


Mas  cos²(1 − 2x)  = 1 − sen²(1 − 2x):

     $\displaystyle=-\,\frac{1}{2}\int\mathrm{sen^3}(1-2x)\cdot \big[1-\mathrm{sen^2}(1-2x)\big]\cdot \big[\!-2\cos(1-2x)\big]\,dx$


Faça a seguinte substituição:

     $\mathrm{sen}(1-2x)=u\quad\Rightarrow\quad -2\cos(1-2x)\,dx=du$


e a integral fica

     $\displaystyle=-\,\frac{1}{2}\int u^3\cdot [1-u^2]\,du\\\\\\ =-\,\frac{1}{2}\int [u^3-u^3\cdot u^2]\,du\\\\\\ =-\,\frac{1}{2}\int [u^3-u^5]\,du\\\\\\ =-\,\frac{1}{2}\cdot \left[\frac{u^{3+1}}{3+1}-\frac{u^{5+1}}{5+1}\right]+C\\\\\\ =-\,\frac{1}{2}\cdot \left[\frac{u^4}{4}-\frac{u^6}{6}\right]+C\\\\\\ =-\,\frac{u^4}{8}+\frac{u^6}{12}+C$

     $=-\,\dfrac{\mathrm{sen}^4(1-2x)}{8}+\dfrac{\mathrm{sen}^6(1-2x)}{12}+C$    <————    esta é uma resposta possível.


Bons estudos! :-)