# Math Help - Integrals Nightmare

1. ## Integrals Nightmare

I continue to struggle with these integrals...
It has given me quite a headache...

Here are the two problems I'm stuck in:

$
\int_1^{-1}\frac{3}{7-5x}dx
$

$
\int_0^\frac{2\pi}{3} tan(\frac{x}{2})dx

$

What confuses me the most are the numbers on the Integral sign...

I continue to struggle with these integrals...
It has given me quite a headache...

Here are the two problems I'm stuck in:

$
\int_1^{-1}\frac{3}{7-5x}dx
$

$
\int_0^\frac{2\pi}{3} tan(\frac{x}{2})dx

$

What confuses me the most are the numbers on the Integral sign...

These are definite integrals. They are evaluated by applying the FTC [Fundamental Theorem of Calculus] which states that

$\int_a^b f(x)\,dx=F(b)-F(a)$

where $F(x)$ is the anti-derivative.

For the first one:

$\int_1^{-1}\frac{3}{7-5x}\,dx=-\int_{-1}^1\frac{3}{7-5x}\,dx$

Make the u substitution $u=7-5x\implies \,du=-5\,dx$

We can also change the limits of integration:

$u(1)=7-5(1)=2$ and $u(-1)=7-5(-1)=12$

So we have $\tfrac{3}{5}\int_{12}^2\frac{\,du}{u}=-\tfrac{3}{5}\int_2^{12}\frac{\,du}{u}$

Can you take it from here?

You already asked the second one earlier. I gave you a hint for that one here.

--Chris

3. I see...

I'm just a little confused with the whole 'anti-derivatives' part. We just started doing this whole Integral thing and I couldn't understand those numbers on the Integral sign.

So at:
$
\tfrac{3}{5}\int_{12}^2\frac{\,du}{u}=-\tfrac{3}{5}\int_2^{12}\frac{\,du}{u}
$

I need to put in the numbers from the Integral sign to the U's?

For example,
$
-\tfrac{3}{5}\int_2^{12}\frac{\,du}{u}=-\tfrac{3}{5}\int_2^{12}\frac{\,d(12)}{(12)}
$

I see...

I'm just a little confused with the whole 'anti-derivatives' part. We just started doing this whole Integral thing and I couldn't understand those numbers on the Integral sign.

So at:
$
\tfrac{3}{5}\int_{12}^2\frac{\,du}{u}=-\tfrac{3}{5}\int_2^{12}\frac{\,du}{u}
$

I need to put in the numbers from the Integral sign to the U's?

For example,
$
-\tfrac{3}{5}\int_2^{12}\frac{\,du}{u}=-\tfrac{3}{5}\int_2^{12}\frac{\,d(12)}{(12)}
$
No...you plug in the numbers after you integrate.

When you find the anti-derivative, ask yourself which function has a derivative of ____.

In this case, we're integrating $\int\frac{\,du}{u}$

Now, what function has a derivative of $\frac{1}{u}$? Its the natural log function: $\ln u$

So, $\int\frac{\,du}{u}=\ln u+C$

However, since we are dealing with definite integration, we don't need this constant $C$

Now, apply the FTC to evaluate the definite integral:

$-\tfrac{3}{5}\int_2^{12}\frac{\,du}{u}=-\tfrac{3}{5}\left.\bigg[\ln u\bigg]\right|_2^{12}=-\tfrac{3}{5}\left[\ln(12)-\ln(2)\right]=\color{red}\boxed{-\tfrac{3}{5}\ln\left(6\right)}$

Can you try to do the other one?

--Chris

5. Okay, I'm going to try the second.

This is where I'm at so far:

I got my formula ready =>
$
\int_a^b f(x)\,dx=F(b)-F(a)
$

Problem =>
$
\int_0^\frac{2\pi}{3} tan(\frac{x}{2})dx
$

---

$
\int_0^\frac{2\pi}{3} tan(\frac{x}{2})dx= -\int_\frac{2\pi}{3}^0 tan(\frac{x}{2})dx
$

$u = \frac{x}{2}\implies du = tan(u)dx$ The substitution part confuses me... How would you do this?

6. u = x/2
du = 1/2 dx
2du = dx.

Now plug in u for x/2 and du for dx.

7. Hi Terr!

So you are saying that it should be:

$
u = \frac{x}{2}\implies du = \frac{1}{2}(dx)\implies 2du=dx
$

I'm still confused with the number plugging... I was looking at how Chris was doing the limits of integration and I understand how he gets it...
I think I just need a little more of a push. I believe I can solve this problem on my own but I'm just lost.

8. Do the integral first, and don't worry about the limits of integration. The integral seems pretty easy, since you already substituted in and all you have to do is to integrate tan(u). After you finish that integration, you should get some function, say f(x). This is where you start worrying about those little numbers. Plug them in, subtraction the bottom from the top using that algorithm and you should get your answer.

9. I'm honestly a failure...

It's amazing that you guys can calculate all this. I feel like a dummy in class when I try to understand what is going on... -_-

I'm going to go find a tutor for Calculus on this stuff...

I'm sorry, Terr, but I didn't really understand what you wanted me to do. Yeah, my stupidity is amazing. Haha.

10. You have the \int 2tan(u) du. The integral of tangent is -ln(cos(u)), so you have -2ln(cos(x/2)) + C. Now plug in the values that are given. 2pi/3 and 0.

-2ln(cos(pi/3)) + 2ln(cos(0))
-2ln(1/2) + 2ln(1)
-2ln(1/2) + 0
ln(2^((-1)(-2))) = ln(4).
The C cancels itself out. My math is bad, so I probably made many errors inbetween, check it for yourself.

$
\int_0^\frac{2\pi}{3} tan(\frac{x}{2})dx= -\int_\frac{2\pi}{3}^0 tan(\frac{x}{2})dx
$
I should also clarify something.

The proper form of a definite integral is $\int_a^b f(x)\,dx$, where $a

So we see what $\int_0^\frac{2\pi}{3} \tan\left(\frac{x}{2}\right)\,dx$ is in its proper form, since $0<\frac{2\pi}{3}$

However, what if the limits were in reverse order, like in $\int_{\frac{2\pi}{3}}^0 \tan\left(\frac{x}{2}\right)\,dx$ ??

Then we must swap the limits. But doing so will make the integral negative.

So $\int_{\frac{2\pi}{3}}^0 \tan\left(\frac{x}{2}\right)\,dx=-\int_0^\frac{2\pi}{3} \tan\left(\frac{x}{2}\right)\,dx$

So you only swap limits if the upper limit is smaller than the lower limit.

Does this make sense?

--Chris