Answer to Question #178348 in Calculus for Phyroe

Question

Integration Procedures(Itevrations by Parts)

∫ x³ cos 5x dx

Answer 1

The basic formula is

$\displaystyle\int u\ dv=uv-\int v\ du$

The formula may be applied again if necessary

$\displaystyle\int x^3 \cos 5x\ dx=$

denote $u=x^3$ and $dv=\cos 5x\ dx$

then $du=3x^2 dx$ and $v=\dfrac{1}{5}\sin5x$

$\displaystyle x^3\cdot\dfrac{1}{5}\sin5x-\int \dfrac{1}{5}\sin5x\cdot3x^2 dx=$

simplify

$\displaystyle \dfrac{1}{5}x^3\sin5x-\dfrac{3}{5}\int \sin5x\cdot x^2 dx=$

$\displaystyle \dfrac{1}{5}x^3\sin5x-\dfrac{3}{5}\int x^2\cdot\sin5x dx=$

integate by parts again

denote $u=x^2$ and $dv=\sin5x\ dx$

then $du=2x dx$ and $v=-\dfrac{1}{5}\cos5x$

$\displaystyle \dfrac{1}{5}x^3\sin5x-\dfrac{3}{5}\left(x^2\cdot\left(-\dfrac{1}{5}\cos5x\right)-\int \left(-\dfrac{1}{5}\cos5x\right)\cdot2x dx\right)=$

$\displaystyle \dfrac{1}{5}x^3\sin5x-\dfrac{3}{5}\left(-\dfrac{1}{5}x^2\cos5x+\dfrac{2}{5}\int \cos5x\cdot x dx\right)=$

$\displaystyle \dfrac{1}{5}x^3\sin5x+\dfrac{3}{25}x^2\cos5x-\dfrac{6}{25}\int x\cdot\cos5x dx=$

integrate by parts again

denote $u=x$ and $dv=\cos 5x\ dx$

then $du=dx$ and $v=\dfrac{1}{5}\sin5x$

$\displaystyle \dfrac{1}{5}x^3\sin5x+\dfrac{3}{25}x^2\cos5x-\dfrac{6}{25}\left(x\cdot\dfrac{1}{5}\sin5x-\int \dfrac{1}{5}\sin5x\cdot dx\right)=$

$\displaystyle \dfrac{1}{5}x^3\sin5x+\dfrac{3}{25}x^2\cos5x-\dfrac{6}{25}\left(\dfrac{1}{5}x\sin5x- \dfrac{1}{5}\int\sin5x\ dx\right)=$

$\displaystyle \dfrac{1}{5}x^3\sin5x+\dfrac{3}{25}x^2\cos5x-\dfrac{6}{125}x\sin5x+\dfrac{6}{125}\int\sin5x\ dx=$

$\displaystyle \dfrac{1}{5}x^3\sin5x+\dfrac{3}{25}x^2\cos5x-\dfrac{6}{125}x\sin5x+\dfrac{6}{125}\left(-\dfrac{1}{5}\cos5x\right)+c=$

$\displaystyle \dfrac{1}{5}x^3\sin5x+\dfrac{3}{25}x^2\cos5x-\dfrac{6}{125}x\sin5x-\dfrac{6}{625}\cos5x+c$