Question

1) calcule cada integral
A) ∫x²cosxdx

b) ∫raiz cubica de x² lnxdx

Answer 1

Utilizaremos integração por partes

a)

$\displaystyle\int x^{2}cos(x)dx\\\\\\u=x^{2}~~~~~~~~~~~~~~\rightarrow~~du=2xdx\\dv=cos(x)dx~~~~\rightarrow~~~ v=sen(x)$

Integrando por partes, temos:

$\displaystyle\int udv=uv-\int vdu\\\\\\\int x^{2}cos(x)dx=x^{2}\cdot sen(x)-\int sen(x)\cdot 2xdx\\\\\\\int x^{2}cos(x)dx=x^{2}\cdot sen(X)-2\int xsen(x)dx$

Vamos usar integração por partes novamente, para integrar xsen(x)

$a=x~~~~~~~~~~~~~~~\rightarrow~~da=dx\\db=sen(x)dx~~~\rightarrow~~~b=-cos(x)\\\\\\\displaystyle\int adb=ab-\int bda\\\\\\\int xsen(x)dx=x\cdot(-cos(x))-\int-cos(x)dx\\\\\\\int xsen(x)dx=-xcos(x)+\int cos(x)dx\\\\\\\boxed{\boxed{\int xsen(x)dx=-xcos(x)+sen(x)+c_{1}}}$

Substituindo na expressão, temos

$\displaystyle\int x^{2}cos(x)dx=x^{2}\cdot sen(x)-2\int xsen(x)dx\\\\\\\int x^{2}cos(x)dx=x^{2}sen(x)-2\left(-xcos(x)+sen(x)+c_{1}\right)\\\\\\\int x^{2}cos(x)dx=x^{2}sen(x)+2xcos(x)-2sen(x)+2c_{1}\\\\\\\boxed{\boxed{\int x^{2}cos(x)dx=x^{2}sen(x)+2xcos(x)-2sen(x)+C}}$
__________________________

b)

$\displaystyle\int\sqrt[3]{x^{2}}ln(x)dx=\int x^{2/3}ln(x)dx$

Escolhendo u e dv:

$u=ln(x)~~~\rightarrow~~~du=\frac{1}{x}dx\\dv=x^{2/3}dx~~~\therefore~~~v=(\frac{3}{5})x^{5/3}$

Então:

$\displaystyle\int udv=uv-\int vdu\\\\\\\int\sqrt[3]{x^{2}}ln(x)dx=ln(x)\cdot\dfrac{3}{5}x^{5/3}-\int\dfrac{3}{5}x^{5/3}\dfrac{1}{x}dx\\\\\\\int\sqrt[3]{x^{2}}ln(x)dx=\dfrac{3}{5}x^{5/3}ln(x)-\dfrac{3}{5}\int x^{(5/3)-1}dx\\\\\\\int\sqrt[3]{x^{2}}ln(x)dx=\dfrac{3}{5}x^{5/3}ln(x)-\dfrac{3}{5}\cdot\dfrac{x^{(5/3)-1+1}}{(5/3)-1+1}+C\\\\\\\int\sqrt[3]{x^{2}}ln(x)dx=\dfrac{3}{5}x^{5/3}ln(x)-\dfrac{3}{5}\dfrac{x^{5/3}}{(5/3)}+C$

$\displaystyle\int\sqrt[3]{x^{2}}ln(x)dx=\dfrac{3}{5}x^{5/3}ln(x)-\dfrac{3}{5}\dfrac{3}{5}x^{5/3}+C\\\\\\\boxed{\boxed{\int\sqrt[3]{x^{2}}ln(x)dx=\dfrac{3}{5}x^{5/3}\cdot\left(ln(x)-\dfrac{3}{5}\right)+C}}$