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anonymous
 5 years ago
integration of [(x+ pi)^3+ cos^2 (x+3*pi)] dx from (3pi/2) to (pi/2)
anonymous
 5 years ago
integration of [(x+ pi)^3+ cos^2 (x+3*pi)] dx from (3pi/2) to (pi/2)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{\frac{3\pi}{2}}^{\frac{\pi}{2}} [(x+ \pi)^3 + \cos^2(x+3\pi)]dx\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont know why i can't open the equation editor..:(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is this the question?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yup.. the Q isn't confusing you ryt?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let me think about it first :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sure.. take your time.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you 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
 5 years ago
Best ResponseYou've already chosen the best response.0hey it worked! thanx!:)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you 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
 5 years ago
Best ResponseYou've already chosen the best response.0oh 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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