Answer to Question #178348 in Calculus for Phyroe

Question #178348

Integration Procedures(Itevrations by Parts)


∫ x³ cos 5x dx


1
Expert's answer
2021-04-15T07:27:13-0400

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"




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS