Integration (Calc.) question?

Question

How do you integrate sin^2(3x)dx without Euler's formula?

Answer 1

∫(sin^2(3x) dx)

To integrate this, you need to use the half angle identity, which goes as follows:

sin^2(y) = (1/2)(1 - cos(2y))

In this case, y = 3x, so we get

∫( (1/2)(1 - cos(2*3x)) dx )

Factor out the (1/2) and simplify.

(1/2) ∫ ( ( 1 - cos(6x) ) dx )

Now we can integrate term by term. Remember that the integral of cos(kx) is equal to (1/k)sin(kx) for some constant k.

(1/2) (x - (1/6)sin(6x) ) + C

Distribute the (1/2),

(1/2)x - (1/12)sin(6x) + C

Answer 2

∫(sin^2(3x) dx)
To integrate this, you need to use the half angle identity:
sin^2(y) = (1/2)(1 - cos(2y))
In this case, y = 3x, so we get
∫( (1/2)(1 - cos(2*3x)) dx )
Factor out the (1/2) and simplify.
(1/2) ∫ ( ( 1 - cos(6x) ) dx )
Now we can integrate term by term. Remember that the integral of cos(kx) is equal to (1/k)sin(kx) for some constant k.
(1/2) (x - (1/6)sin(6x) ) + C
Distribute the (1/2),
(1/2)x - (1/12)sin(6x) + C

Answer 3

First, let's make it simpler:
Let u = 3x, x = u/3, dx = du/3.
So now we have to compute
1/3 ∫ sin² u du.
To do the last integral, note that
cos² u+ sin² u = 1
cos²u - sin² u = cos 2u
Subtracting, we get
sin² u = (1- cos 2u)/2.
So the integral becomes
1/6*∫ (1 - cos 2u) du =
1/6*[u - sin 2u/2] = 1/6[3x - cos 6x] + C.

Answer 4

substitute u = 3x dx
du = 3 dx
so it would be
1/3 int( sin^2(u) du)
use trig substitute or the back of you book for the integral of sin^2(u)
the integral is 1/2u - 1/4sin(2u)
substitute back what you took u as
than your integral of sin^2(3x) is
1/3(1/2(3x) - 1/4sin(2(3x)) + C

trigonometric integral

use sin^2(x) as 1/2(1-cos(2x))
this is the half-angle identity
integral 1/2(1-cos(2x)

Answer 5

integral sin^2(3x)dx =
integral (1+cos6x)/2 dx
= x/2 + sin6x/12 + c

Answer 6

You have always sin ^2(3x) +cos^2(3x)=1
and -sin^2(3x) +cos^2(3x) = cos 6x so substracting

sin^2(3x) = 1/2 (1-cos 6x)
The integral is 1/2(x+1/6*sin 6x) +C

Answer 7

int(cos(x)/(2+sin(x)^2), x) = (a million/2)*sqrt(2)*arctan((a million/2)*sin(x)*sqrt(2... + C int(x^2*e^(3*x), x) = (a million/27)*(2-6*x*ln(e)+9*x^2*ln(e)^2)*e^(3*... + C > int(x*e^((-2)*x), x) = - (a million/4)*(a million+2*x*ln(e))*e^(-2*x)/ln(e)^2 + C

Answer 8

what a fool u r