Integrating Tricky Functions: Solving and Proving Integrals

[MathematicalPhysicist]

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)

 

[neutrino]

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.

 

[Max Eilerson]

substitute u = 1 + sin x. Are you sure the integral is equal to 4, not -4?

 

[neutrino]

But the plot (area) of the function sqrt(1+sin(x)) from 0 to pi seems to be non-zero!

 

[Max Eilerson]

It's non-zero.

 

[MathematicalPhysicist]

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.

 

[Max Eilerson]

complete the square.

 

[MathematicalPhysicist]

you mean something like this: u^2+u+4=(u-2)^2+5u
i still don't get an appropiate term to integrate.

 

[neutrino]

More like (u +0.5)^2 + 15/4

 

[MathematicalPhysicist]

ok, thanks.
btw, what about the second integral does it equal zero or it really does equal 4?

 

[Max Eilerson]

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.

 

[neutrino]

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.

 

[Max Eilerson]

Btw, you will need to know what the derivative of the inverse tangent is.

 

[Max Eilerson]

Oops, I'm missing/added a minus somewhere. Didn't have the common senese to think about the graph :).

 

[MathematicalPhysicist]

wait a minute, then integral does converge to 4, care to explain how, where did i get it wrong?

 

[dextercioby]

For the integral try the sub

$$\tan\frac{x}{2}=t$$

Daniel.

 

[MathematicalPhysicist]

but what's wrong with the substitution that I am given a hint to use here?
i.e
sinx=u?

 

[HallsofIvy]

If you let u= sin(x) then du= cos(x)dx so your integral will involve something like $\frac{du}{cos(x)}$ (with the cos(x) converted to u of course) but $cos(\pi/2)= 0$ so that is not a valid substitution.

 

[NateTG]

$$\int_{0}^{\pi}\sqrt{1+sinx}dx=4$$
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)

You need to be careful about which square root you're using: The integrand should be:
$$\frac{\sqrt{1+u}}{\pm \sqrt{1-u^2}}$$
The trick is that you'll be using one square root from $x=0$ to $x=\frac{\pi}{2}$ and the other root from $x=\frac{\pi}{2}$ to $x=0$.

 

[MathematicalPhysicist]

ok, i understand the trick here, i haven't seen this point x=pi/2 as a "bad" point, thank you for the pointers.