Question

integration of [(x+ pi)^3+ cos^2 (x+3*pi)] dx from (-3pi/2) to (-pi/2)

Answer 1

i

Answer 2

\[\int\limits_{\frac{-3\pi}{2}}^{\frac{-\pi}{2}} [(x+ \pi)^3 + \cos^2(x+3\pi)]dx\]?

Answer 3

i dont know why i can't open the equation editor..:-(

Answer 4

is this the question?

Answer 5

yup.. the Q isn't confusing you ryt?

Answer 6

let me think about it first :)

Answer 7

sure.. take your time.

Answer 8

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 :)

Answer 9

hey it worked! thanx!:-)

Answer 10

np ^_^

Answer 11

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 :|

Answer 12

didnt*

Answer 13

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 \[\pi/2\]