## computation of integrals.

i need to solve/prove the next two integrals:
$$\int\frac{dx}{u^2+u+4}$$
and i need to show that:
$$\int_{0}^{\pi}\sqrt{1+sinx}dx=4$$
the problem is that i have a clue to substitute u=sinx and then sin(pi)=0=sin0 so the integral should be equal zero, is it not?
ofcourse the integrand becomes: sqrt(1+u)/sqrt(1-u^2)
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 For the first you can use the method of partial fractions. You're right about the second. With the given limits, the integral is equal to zero.
 substitute u = 1 + sin x. Are you sure the integral is equal to 4, not -4?

## computation of integrals.

But the plot (area) of the function sqrt(1+sin(x)) from 0 to pi seems to be non-zero!
 It's non-zero.
 neutrino, how would i use partail fractions here? i mean i need to decompose u^2+u+4 into a product of terms, but i have complex roots here.
 complete the square.
 you mean something like this: u^2+u+4=(u-2)^2+5u i still dont get an appropiate term to integrate.
 More like (u +0.5)^2 + 15/4
 ok, thanks. btw, what about the second integral does it equal zero or it really does equal 4?
 u^2+u+4= (u + 0.5)^2 + 15/4. edit: too slow, the second integral should equal minus -4, I guess they're defining is it as area so you just need the modulus.
 Not -4. I just put the function through the Integrator and substituted the values, and I got 4. This graph is completely above the x-axis.
 Btw, you will need to know what the derivative of the inverse tangent is.
 Oops, I'm missing/added a minus somewhere. Didn't have the common senese to think about the graph :).
 wait a minute, then integral does converge to 4, care to explain how, where did i get it wrong?
 Blog Entries: 9 Recognitions: Homework Help Science Advisor For the integral try the sub $$\tan\frac{x}{2}=t$$ Daniel.
 but what's wrong with the substitution that im given a hint to use here? i.e sinx=u?