Hello, I've been struggling with three anti-derivative problems in calculus, I will present them here along with my attempted answers.

1) Integral of (x^5-x^2)((x^3+1)^1/3)dx

my answer:

u = x^3 +1
du = 3x^2dx
dx = du/3x^2

((x^5-x^2)/(3x^2)) MULTIPLIED BY (((3((x^3+1)^4/3))/4)/(3x^2))

2) Integral of 5sec(5x)tan(5x)dx

Alt: (1/5cos(5x)) MULTIPLIED BY (sin(5x)/cos(5x)) MULTIPLIED BY dx

u = cos(5x)
du = -sin(5x) dx
dx = du/-sin(5x)

(1/5cos(5x)) MULTIPLIED BY (sin(5x)/cos(5x)) MULTIPLIED BY dx

My answer was 25sec(5x)+C

Let me know if there is any more information I can provide!

What is going to save you is all that algebra (and sometimes trig) that you learned up to now.
The other thing is when you substitute u = f(x) and du = f'(x) dx, you need to substitute all the x's out for u-terms. We are in total x-eradication mode. Hunt and destroy. But do it with the tools of algebra where ∫ and d_ are just funny little markers.

∫(x^5-x^2)(x^3+1)^(1/3) dx

Guess u = (x^3+1) since that is clearly the ugliest part of this integrand
Then du = 3 x^2 dx -- We need to find this factor so we can swap in the left side of the equation for the right side.

1 = (1/3)(3) = (-1)(-1) -- useful!
And anything times 1 is left unchanged.
0 = 1 - 1 -- Also useful!
And anything plus zero is left unchanged.

= ∫(1/3)(x^3-1)(x^3+1)^(1/3) (3 x^2 ) dx
= ∫(1/3)(x^3+1-1-1)(u)^(1/3) du
Hey, we found another u!

=∫(1/3)(u-2)(u)^(1/3) du
Now we split
= (1/3)∫(u)^(4/3) du - (2/3) ∫(u)^(1/3) du

We rock!

= (1/3)(3/7) u^(7/3) - (2/3)(3/4) u^(4/3) + C
= (2/14) u u^(4/3) - (7/14) u^(4/3) + C
= (1/14) u^(4/3) ( 2 u - 7 ) + C

But u is not useful to our Prof.

= (1/14) (x^3+1)^(4/3) (2(x^3+1) - 7) + C
= (1/14) (x^3+1)^(4/3) (2 x^3 - 5) + C

There are a couple of ways to write this in a "nice" way, but all of them are trivial to express from this point, so we are done.

∫ 5 sec(5x) tan(5x) dx
But sec() is 1/cos() and tan() = sin()/cos()

So this is
= ∫ 5 sin(5x) ( cos(5x) )^-2 dx

Obviously u = cos(5x) is the ugliest part of the integrand.
Then du = -5 sin(5x) dx

Can you find a factor of -5 sin(5x) in the integrand?

So go for it! And show us how you do.


Was there a third problem? -- Also, thanks for correctly using the Homework Help section.

Hello rpenner!

And thanks for your help so far.

I'm currently reviewing problem 1 and am starting to understand the algebraic gymnastics displayed here. Still struggling though. I'd say the biggest point of origin for my problems is that when the material is presented, it is watered down (or flat out incorrect) and when the same material appears on an assignment, it is compounded with more complicated concepts.

So here is an example straight from my book...I am not sure how we make the jump from talking about e, to u, to tan^-1:

---EDIT---

Well I was going to link directly to the image but linking is not an available option to new members.

---

Integral((dx)/e^x+e^-x) MULTIPLIED BY (e^x)/(e^x)
Integral(((e^x)dx)/(e^2x +1)) U EQUALS e^x; U^2 EQUALS e^2x; du = (e^x)dx
Integral((du)/((u^2)+1))
INTEGRATE WITH RESPECT TO U = (tan^-1)u+C
REPLACE U WITH e^x
(tan^-1)(e^x)+C

Now...how did we go from talking about e, to u, to inverse tangent?

That's just the anti-derivative of 1/(x^2 +1)

http://mathworld.wolfram.com/InverseTangent.html

And you can demonstrate that with

http://www.themathpage.com/acalc/inverse-trig.htm

∫ dx/(e^x + e^(-x))
= ∫ 1/(e^x + e^(-x)) dx

u = e^x
du = e^x dx

So we need to factor out the du if this is going to work. - 1 = (e^x)(1/e^x)

= ∫ (e^x)(1/e^x)1/(e^x + e^(-x)) dx
= ∫ (1/e^x)1/(e^x + e^(-x)) du
= ∫ (1/e^x)1/(e^x + e^(-x)) du
= ∫ 1/((e^x)^2 + (e^(-x))(e^x)) du
= ∫ 1/((e^x)^2 + 1) du
= ∫ 1/(u^2 + 1) du
= arctan(u) + C
= arctan(e^x) + C

Rpenner, Thank you for showing me where to look! I've got the tools to figure some of this stuff out now.

So I'm following a lower level example and am suspicious of the answer that is presented...

...Here is the problem:

Find Integral((sin(x)+1)^7)(dx))

u = sin(x) + 1
du = cos(x)dx
dx = du/cos(x)

The answer is:

(sin(x)+1)^8 ALL OVER 8 PLUS C

...does this ignore the implied 1/cos(x) outside of the integral?

ALSO! Do you have any documentation on splitting the integral? This is a technique that was left out of the lecture.

You can't take 1/cos(x) "outside" of the integral -- it is not a constant like "3", but is dependent on x.

Can you say where (textbook, author, copyright year, and page) this example is given?

If you differentiate (sin(x)+1)^8/8 + C, the answer is (sin(x)+1)^7 cos(x), so obviously there is mistake somewhere.

--

On splitting the integral -- you can do this by replacing the two parts inside the integral with two integrals with a plus sign.

You can do it "vertically" with the limits of integration:

∫ (from -a to b ) f(x) dx = ∫ (from -a to 0) f(x) dx + ∫ (from 0 to b ) f(x) dx (-a < 0 < b )

∫ (from a to b ) f(x) dx = ∫ (from a to c) f(x) dx + ∫ (from c to b ) f(x) dx (a < c < b )

∫ (from a to b ) f(x) dx = ∫ (from a to d) f(x) dx - ∫ (from b to d) f(x) dx ( a < b < d )

[ because ∫ (from a to b ) f(x) dx + ∫ (from b to d) f(x) dx = ∫ (from a to d) f(x) dx ]


Or you can do it horizontally:

∫ f(x) + g(x) dx = ∫ f(x) dx + ∫ g(x) dx



So ∫ (sin(x))^3 dx = ∫ (sin(x))^2 sin(x) dx = ∫ (1 - (cos(x))^2) sin(x) dx = ∫ sin(x) - (cos(x))^2 sin (x) dx

Time to split into ∫ sin(x) dx + ∫ - (cos(x))^2 sin (x) dx
∫ sin(x) dx = -cos(x) + C
∫ - (cos(x))^2 sin (x) dx = ∫ (u)^2 du = u^3/3 + C = (cos(x)^3)/3 + C
So
∫ (sin(x))^3 dx = (cos(x))^3/3 - cos(x) + C

Test: d ( (cos(x))^3/3 - cos(x) + C ) / dx = (cos(x))^2 (-sin(x)) + sin(x) = sin(x)(1 - (cos(x))^2 ) = (sin(x))^3


And this works when parts of the integrand are connected by addition (or subtraction) signs.

If they are connected by multiplication, a much trickier subject called integration by parts is needed, but that will be the subject of an entire lecture, not just tacked on at the end of another.

I'm currently parsing the information you sent me just now; however, I have a screen shot of the example. Provide me a way to send you a .png and I will send you the example involving trigonometry.