anonymous
  • anonymous
integration of [(x+ pi)^3+ cos^2 (x+3*pi)] dx from (-3pi/2) to (-pi/2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
i
anonymous
  • anonymous
\[\int\limits_{\frac{-3\pi}{2}}^{\frac{-\pi}{2}} [(x+ \pi)^3 + \cos^2(x+3\pi)]dx\]?
anonymous
  • anonymous
i dont know why i can't open the equation editor..:-(

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anonymous
  • anonymous
is this the question?
anonymous
  • anonymous
yup.. the Q isn't confusing you ryt?
anonymous
  • anonymous
let me think about it first :)
anonymous
  • anonymous
sure.. take your time.
anonymous
  • anonymous
you can cut it down into the following:\[\int\limits_{\frac{-3\pi}{2}}^{\frac{-\pi}{2}}(x+\pi)^3 dx +\int\limits_{\frac{-3\pi}{2}}^{\frac{-\pi}{2}}\cos^2(x+3\pi)dx \] for the first part, expand the polynomial and integrate normally, and as for the second part, let u = x+3pi and integrate using the half angle formula for cos^2 u, which is in this case : \[\cos^2u = \frac{1}{2}(1+\cos2u)\] give it a try :)
anonymous
  • anonymous
hey it worked! thanx!:-)
anonymous
  • anonymous
np ^_^
anonymous
  • anonymous
you did even need to expand the first bit\[\int\limits_{}^{} (ax+b)^n = (ax+b)^{n+1} / [a(n+1)] +C\] I cant manage to get a straight horizontal bar for fractions when working with the equation editor :|
anonymous
  • anonymous
didnt*
anonymous
  • anonymous
oh that expansion part jus worked out easy using a property of definite integrals.. \[\int\limits_{a}^{b} x dx = (b+a - x)dx\] plugging in the values of the question the whole thing comes out negative. thus adding them would yield zero. solving the trig part yields \[\pi/2\]

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