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

1. anonymous

i

2. anonymous

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

3. anonymous

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

4. anonymous

is this the question?

5. anonymous

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

6. anonymous

let me think about it first :)

7. anonymous

8. 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 :)

9. anonymous

hey it worked! thanx!:-)

10. anonymous

np ^_^

11. 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 :|

12. anonymous

didnt*

13. 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 $\pi/2$