# Thread: Help with these integrals:

1. ## Help with these integrals:

1.)

Is it necessary to use integration by parts here? cause it wasnt discussed in this lesson yet

2.)

and thanks a kazillions

2. Originally Posted by ^_^Engineer_Adam^_^
1.)

Is it necessary to use integration by parts here? cause it wasnt discussed in this lesson yet

2.)

and thanks a kazillions
The second integratoin has me stumped right now. I'll try coming back to it later, but for now, I can help you with the first integration:

1. INT(Sqrt(3 - 2x)x^2) dx
This can be solved without integration by parts. You might find it hard to believe, but we can actually solve this by u substitution:

Let u = 3 - 2x ... Why would I do that? You'll see .
du = -2dx --> dx = -du/2
Before I go on, let's solve the u = 3 - 2x for x (we'll need this in a second): x = 1/2(3 - u)

Ok, now watch my slight of hand as I do some math-magic on this integration...

INT(Sqrt(u)x^2) -du/2 ... Now you see an x^2,
-1/2 INT(Sqrt(u)[1/2(3 - u)]^2) du ... Now you don't. ... Guess where it went. (Hint: x = 1/2(3 - u))

From here, I think you can solve it the rest of the way, just multiply out the [1/2(3 - u)]^2 then multiply this to the Sqrt(u).

3. Originally Posted by ^_^Engineer_Adam^_^
1.)

Is it necessary to use integration by parts here? cause it wasnt discussed in this lesson yet

2.)

and thanks a kazillions
This second integration throws a giant curve ball at you. I've figured out how to do it, it's just that the way to do it is a little tricky.

INT (tan(2x) + cot(2x))^2 dx ... let's turn these into sines and cosines
INT (sin(2x)/cos(2x) + cos(2x)/sin(2x))^2 ... now, get common denominators to add the fractions
INT [((sin(2x))^2 + (cos(2x))^2)/(sin(2x)cos(2x)]^2 dx
INT 1/((sin2x)^2(cos2x)^2) dx

Now here's the catch, we can use (sin2x)^2 + (cos2x)^2 to replace the 1 in the numerator.

You might be thinking, why use that identity in reverse when we already used it to get rid of (sin2x)^2 + (cos2x)^2?

The reason we had to get rid of the (sin2x)^2 + (cos2x)^2 a few steps before is because that whole term was being raised to the second power. If we had multiplied all that out, the problem would get really messy, but there's nothing wrong with replacing 1 with (sin2x)^2 + (cos2x)^2, so that's what we'll do.

INT ((sin2x)^2 + (cos2x)^2)/((sin2x)^2(cos2x)^2) dx

Now let's split up this fraction again. We should get:

INT 1/(cos2x)^2 + 1/(sin2x)^2 dx
INT (sec2x)^2 + (csc2x)^2 ... This we can integrate.

4. Originally Posted by ecMathGeek
This second integration throws a giant curve ball at you. I've figured out how to do it, it's just that the way to do it is a little tricky.

INT (tan(2x) + cot(2x))^2 dx ... let's turn these into sines and cosines
INT (sin(2x)/cos(2x) + cos(2x)/sin(2x))^2 ... now, get common denominators to add the fractions
INT [((sin(2x))^2 + (cos(2x))^2)/(sin(2x)cos(2x)]^2 dx
INT 1/((sin2x)^2(cos2x)^2) dx

Now here's the catch, we can use (sin2x)^2 + (cos2x)^2 to replace the 1 in the numerator.

You might be thinking, why use that identity in reverse when we already used it to get rid of (sin2x)^2 + (cos2x)^2?

The reason we had to get rid of the (sin2x)^2 + (cos2x)^2 a few steps before is because that whole term was being raised to the second power. If we had multiplied all that out, the problem would get really messy, but there's nothing wrong with replacing 1 with (sin2x)^2 + (cos2x)^2, so that's what we'll do.

INT ((sin2x)^2 + (cos2x)^2)/((sin2x)^2(cos2x)^2) dx

Now let's split up this fraction again. We should get:

INT 1/(cos2x)^2 + 1/(sin2x)^2 dx
INT (sec2x)^2 + (csc2x)^2 ... This we can integrate.
YOU ARE THE MAN ECMATHGEEK!!!!! by the way, what does the ec stand for?

5. Originally Posted by Jhevon
YOU ARE THE MAN ECMATHGEEK!!!!! by the way, what does the ec stand for?
Thank you.

The real reason I have the letters e and c in my handle is kind of lame, so let's just say I really like the numbers e and c.

I use the "ec" with "MathGeek" because "MathGeek" alone is usually taken in most things I try to sign up for, whereas "ecMathGeek" never is.

6. Originally Posted by ^_^Engineer_Adam^_^
1.)

7. For the second one just expand it and then it's done...

8. Originally Posted by Krizalid
For the second one just expand it and then it's done...
... I tried expanding it, it became a giant mess.

Ok, maybe not a giant mess, but it wasn't easy to integrate without some special trig-identity substitutions.

9. ## Second one:

int[(tan(2x)+cot(2x))^2]dx =

int[ (sin(2x)/cos(2x) + cos(2x)/sin(2x))^2]dx =

int[ ((sin(2x))^2 + (cos(2x))^2))/ cos(2x)sin(2x))^2 ] =

int[(1/sin(2x)cos(2x))^2]dx (i used sin(2x)^2 + cos(2x)^2 = 1 identity)

= subtitue 2x=t ==> dx = dt/2

= 1/2int[ (1/costsint)^2] dt (2sinxcosx = sin2x ==> sinxcosx = 1/2sin2x)

= 1/2int[ (2/sin2t)^2 ]dt = 2int[1/sin(2t)^2]dt = -2cot(2t)/2 = -cot4x

i guess you can open -cot4x for the answer they request using identities

10. Thanks a lot rookies

11. Originally Posted by ecMathGeek
... I tried expanding it, it became a giant mess.
I don't think so, look

As desired...

12. Originally Posted by Krizalid
I don't think so, look

As desired...
lol. Nicely done. I was intent on changing them to sines and cosines and doing so made them a giant mess. I did demonstrate that it could be done with sines and cosines but thank you for clarifying that it wasn't needed. I didn't even notice that that could be done.

The second one can be done without switching functions . . .

[tan(2x) + cot(2x)]² dx

[tan(2x) + cot(2x)]² .= .tan²(2x) + 2·tan(2x)·cot(2x) + cot²(2x)

. . . . . . . . . . . . . . .= .tan²(2x) + 2 + cot²(2x)

. . . . . . . . . . . . . . .= .[sec²(2x) - 1] + 2 + [csc²(2x) - 1]

. . . . . . . . . . . . . . .= .sec²(2x) + csc²(2x)

The integral becomes: .
[sec²(2x) + csc²(2x)] dx . = . ½·tan(2x) - ½·cot(2x) + C