How Do You Solve Integrals Using Integration By Parts?

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SUMMARY

This discussion focuses on solving integrals using the method of Integration by Parts, specifically for the integrals $\int x^n e^x dx$ and $\int \sin^n x dx$. The integration by parts formula, $\int fg \, dx = fg - \int gf' \, dx$, is applied to the first integral, leading to a recursive solution that reduces the exponent of $x$ until it reaches zero. For the second integral, the approach varies based on whether $n$ is even or odd, utilizing trigonometric identities and substitutions to simplify the integral.

PREREQUISITES
  • Understanding of Integration by Parts
  • Familiarity with basic integral calculus
  • Knowledge of trigonometric identities
  • Ability to perform substitution in integrals
NEXT STEPS
  • Practice solving integrals using the Integration by Parts method
  • Explore the use of trigonometric identities in integration
  • Learn about recursive integration techniques
  • Study the properties of even and odd functions in calculus
USEFUL FOR

Students studying calculus, particularly those tackling integration techniques, and educators looking for effective methods to teach integration by parts.

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Homework Statement


1.[tex]$\int x^ne^xdx$[/tex]
2.[tex]$\int \sin ^nxdx$[/tex]


Homework Equations


[tex]$ \displaystyle \Large \int fg dx = fg - \int gf' dx$[/tex]


The Attempt at a Solution


1. f=xn
g'=ex
g=ex
f'=nxn-1

then just plug it in the formula? i tried but i don't get the right answer..

2. i have no idea how to even start..
the antiderivative of sinnx is [(sinx)n+1]/(n+1) ?
 
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No, the anti-derivative of sinn(x) is NOT sinn+1(x)/(n+1)!

Integrating [itex]\int x^n e^x dx[/itex] is very easy- but tedious. Let u= xn, dv= ex dx. Then du= n xn-1 dx and v= ex.

[tex]\int x^n e^x dx= x^ne^x- n\int x^{n-1}e^x dx[/tex]
That is the same as you started with but the exponent on x is one less. Repeat n-1 more times until the exponent is 0!

As for (2), how you do that depends on whether n is even or odd. If n is odd, it is easy. If n= 2m+1, then the integral is
[tex]\int sin^{2m+1}(x) dx= \int (sin^2(x))^m sin(x)dx= \int (1- cos^2(x))^msin(x) dx[/tex]
and letting u= cos(x) reduces it to
[tex]-\int (1- u^2)^m du[/tex]

If n is even, use the trig identity [itex]sin^2(x)= (1/2)(1- cos(2x))[/itex] repeatedly until you have reduced to power 1.
 

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