Question

integral de sen ³ x dx

Answer 1

É dada a integral:

$\displaystyle I=\int\sin^3(x)\,dx$

Vamos resolvê-la por partes:

$\displaystyle I=\int\sin^3(x)\,dx=\int\sin^2(x)\cdot\sin(x)\,dx\\\\ u=\sin^2(x)\Longrightarrow du=2\sin(x)\cos(x)\,dx\\ dv=\sin(x)\,dx\Longrightarrow v=-\cos(x)\\\\ I=uv-\int v\,du=-\sin^2(x)\cos(x)-\int(-\cos(x))\cdot2\sin(x)\cos(x)\,dx\\\\ I=-\sin^2(x)\cos(x)+2\int\sin(x)\cos^2(x)\,dx$

Na última expressão, utilizaremos relação $\sin^2(x)+\cos^2(x)=1\Lonrightarrow \cos^2(x)=1-\sin^2(x)$. Assim:

$\displaystyle I=-\sin^2(x)\cos(x)+2\int\sin(x)\cos^2(x)\,dx\\\\ I=-\sin^2(x)\cos(x)+2\int\sin(x)(1-\sin^2(x))\,dx\\\\ I=-\sin^2(x)\cos(x)+2\int\sin(x)\,dx-2\underbrace{\int \sin^3(x)\,dx}_{I}$ 

Note que a integral destacada acima é igual à integral que queremos calcular (no caso, I). Substituindo:

$\displaystyle I=-\sin^2(x)\cos(x)+2\int\sin(x)\,dx-2\underbrace{\int \sin^3(x)\,dx}_{I}\\\\ I=-\sin^2(x)\cos(x)+2\int\sin(x)\,dx-2I\\\\ 3I=-\sin^2(x)\cos(x)+2\int\sin(x)\,dx\\\\ 3I=-\sin^2(x)\cos(x)+2(-\cos(x))\\\\ 3I=-\sin^2(x)\cos(x)-2\cos(x)\\\\ 3I=-(1-\cos^2(x))\cos(x)-2\cos(x)\\\\ 3I=-\cos(x)+\cos^3(x)-2\cos(x)\\\\ 3I=\cos^3(x)-3\cos(x)\\\\ I=\dfrac{1}{3}\cos^3(x)-\cos(x)\\\\ \boxed{\int\sin^3(x)\,dx=\dfrac{1}{3}\cos^3(x)-\cos(x)}$

Answer 2

$\displaystyle\sf\int sen^3(x)~dx=\int sen^2(x)\cdot sen(x)~dx\\\displaystyle\int (1-cos^2(x))\cdot sen(x)~dx\\\sf fac_{\!\!,}a~u=cos(x)\implies du=-sen(x)~dx\\\displaystyle\sf\int(1-cos^2(x))\cdot sen(x)~dx=-\int(1-u^2)du=-u+\dfrac{1}{3}u^3+k\\\displaystyle\sf\int sen^3(x)~dx=-cos(x)+\dfrac{1}{3} ~cos^3(x)+k$