help, urgent maths stuff?

Question

heya, i have a couple of questions
1) i need to know the logic behind ∫cos²3x dx=1/2x + 1/12sin6x + c

2) how do you do ∫10x(power 4)/ (x (power 5) +1)

any help appreciated

Answer 1

Thankyou - you just reminded me how much math I've forgotten

Answer 2

They wrote cos^2 (3x) as a function of cos(6x)

cos^2 (3x) = 1/2 *[cos(6x) + 1]

That's how they got the above answer.

2) 10x^4 dx / (x^5 + 1)
It is very important to write the dx when you're doing these integrals. It means something and it helps a lot.

Use the substitution u = x^5 + 1
du = 5x^4 dx
So your problem becomes 2*du / u
Integrated you get 2ln(u) = 2*ln(x^5 + 1) + constant

You could also write it more elegantly,
ln [C*(x^5 + 1)^2]

Answer 3

Everyone seems to have answered 1) for you .

The second integral is a matter of recognition. If the numerator is the derivative of the denominator then the answer will always be ln(denominator) + c whatever the denominator is.

You can adjust the numerator by a multiplying constant if necessary, so if your 10 had been a 20 then the answer would be 4ln(x^5 + 1) as the derivative has a 5 in it.

Answer 4

So Lisa scores 20/30. you ought to artwork out 25% of the 20 that Lisa scored. So destroy this down... 50% of 20 (a a million/2) = 10 25% of 20 (a million/2 back) = 5 So 25% of 20 = 5 If Maggie scored 25% extra beneficial than Lisa, you upload the 25% of Lisa's score onto 20 which = 25 hence Lisa scores 20/30, Maggie scores 25/30 desire this enables :)

Answer 5

1). cos² 3x + sin² 3x = 1
..... cos² 3x - sin² 3x = cos 6x
Adding,
cos² 3x = ½(1 + cos 6x).
Now you have to integrate
½∫ (1 + cos 6x) dx = x/2 + sin 6x/12 + C.

2). Write the integral as 2 ∫ 5x^4 dx / (x^5+1)
Let u = x^5 + 1, du = 5x^4 dx
and we get
2 ∫ du/u = 2 ln |u| = 2 ln|x^5+1| + C.

Answer 6

1) Using the trig identity cos(2θ) = 2cos²(θ) - 1, you can change this to (cos(2θ) + 1)/2 = cos²(θ), so that cos²(3x) becomes (cos(6x) + 1)/2. This makes it much easier to integrate.

2)
∫10x^4 / (x^5 +1) dx =
∫ 10x^4 (x^5 +1)^-1 dx =
∫ 5x^4 * 2(x^5 +1)^-1 dx =
-∫ [-2(x^5 +1)^-1] * [5x^4] dx =
- (x^5 + 1)^-2 + c

Answer 7

I = ∫ cos² 3x dx
Let u = 3x
du = 3.dx
I = (1/3).∫cos² u du
Now cos 2u = 2 cos²u - 1
2 cos² u = cos 2u + 1
cos² u = (1/2).cos 2u + 1/2
I = (1 / 3).∫(1/2).cos 2u + 1/2 . du
I = (1/12).sin 2u + u / 6 + c
I = (1/12).sin 6x + x / 2 + c

I = ∫10x(4) / ((x^(5) + 1).dx
I = 2.∫ 5x^(4) / (x^(5) + 1)
I = 2.log (x^(5) + 1) + C

Answer 8

For question 2 try substituting

y=(x^5+1)
dy=5x^4dx

Answer 9

use your brain not technology to do it 4 u !

Answer 10

Try this site:

http://www.math.com/