Question

Alguem me ajuda pf??
calcular as primitivas a seguir aplicando a regra de integração por partes:

Answer 1

$\boxed{\begin{array}{l}\underline{\rm Integrac_{\!\!,}\tilde ao~por~partes}\\\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf \int u\cdot dv=u\cdot v-\int v\cdot du}}}}\\\underline{\rm Para~resolver~esta~quest\tilde ao~vamos~usar}\\\underline{\rm o~macete~do~ILATE}\\\sf I\longrightarrow inversa~trigonom\acute etrica\\\sf L\longrightarrow\ell ogaritmo\\\sf A\longrightarrow artim\acute etica\\\sf T\longrightarrow trigonom\acute etrica\\\sf E\longrightarrow exponencial\end{array}}$

$\large\boxed{\begin{array}{l}\tt vale~lembrar~que~isto~se~trata\\\tt de~um~crit\acute erio~para~escolha~do~''u'\\\tt visando~facilitar~o~c\acute alculo~da~integral\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\sf\int (x+1)\,cos(2x)~dx\\\underline{\rm fac_{\!\!,}a}\\\sf u=x+1\implies du=dx\\\sf dv=cos(2x)~dx\implies v=\dfrac{1}{2}\,sen(2x)\\\displaystyle\sf\int (x+1)\,cos(2x)\,dx=(x+1)\cdot\dfrac{1}{2}sen(2x)-\dfrac{1}{2}\int sen(2x) dx\\\displaystyle\sf\int (x+1)\,cos(2x)\,dx=\dfrac{1}{2}(x+1)sen(2x)+\dfrac{1}{4} cos(2x)+k\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf\int e^{2x}\,sen(x)\,dx\\\underline{\rm fac_{\!\!,}a}\\\sf u=sen(x)\implies du=cos(x)dx\\\sf dv=e^{2x}dx\implies v=\dfrac{1}{2}e^{2x}\\\displaystyle\sf\int e^{2x}\,sen(x)\,dx=\dfrac{1}{2}e^{2x}sen(x)-\dfrac{1}{2}\int e^{2x}\,cos(x)\,dx\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf \int e^{2x}\,cos(x)\,dx\\\underline{\rm fac_{\!\!,}a}\\\sf u_1= cos(x)\implies du=-sen(x)dx\\\sf dv_1= e^{2x}dx\implies v_1=\dfrac{1}{2}e^{2x}\\\displaystyle\sf\int e^{2x}\,cos(x)\,dx=\dfrac{1}{2}e^{2x}cos(x)+\dfrac{1}{2}\int e^{2x}sen(x)\,dx\\\underline{\rm vamos~substituir~na~integral~anterior:}\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf\int e^{2x}\,sen(x)\,dx=\dfrac{1}{2}e^{2x}sen(x)-\dfrac{1}{2}\bigg[\dfrac{1}{2}e^{2x}cos(x)+\dfrac{1}{2}\int e^{2x}sen(x)dx\bigg]\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\sf\int e^{2x}sen(x)\,dx=\dfrac{1}{2}e^{2x}sen(x)-\dfrac{1}{4}e^{2x}\,cos(x)-\dfrac{1}{4}\int e^{2x}\,sen(x)\,dx\\\displaystyle\sf\int e^{2x}\,sen(x)\,dx+\dfrac{1}{4}\int e^{2x}\,sen(x)\,dx=\dfrac{1}{2}e^{2x}\bigg[sen(x)-\dfrac{1}{2}cos(x)\bigg] \\\displaystyle\sf\dfrac{5}{4}\int e^{2x}\,sen(x)\,dx=\dfrac{1}{2}e^{2x}\bigg[ sen(x)-\dfrac{1}{2}cos(x)\bigg]\end{array}}$

$\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\int e^{2x}\,sen(x)\,dx=\dfrac{2}{5}e^{2x}\bigg[sen(x)-\dfrac{1}{2}cos(x)\bigg]+k}}}}$