# Find the integral of 3cos2x-2sinx

## Just goes to show you how important brackets are when typing

in the required format on this webpage.

as 3cos(2x), and his answer is correct according to that

as 3cos^2 (x), and my answer reflects that way of looking at it.

Make sure that you make it clear which way.

∫ [ 3 cos ( 2 x ) - 2 sin x ] dx

then

∫ cos ( 2 x ) dx

Substitute:

u = 2 x

du = 2 dx

du / 2 = dx

dx = du / 2

∫ cos ( 2 x ) dx = ∫ cos u ∙ du / 2 = 1 / 2 ∫ cos u ∙ du = 1 / 2 sin u = 1 / 2 sin ( 2 x )

∫ sin x dx = - cos x

∫ [ 3 cos ( 2 x ) - 2 sin x ] dx =

3 ∫ cos ( 2 x ) dx - 2 ∫ sin x dx = 3 / 2 sin ( 2 x ) - 2 ( - cos x ) + C =

3 / 2 sin ( 2 x ) + 2 cos x + C

## Oh boy, we've got ourselves an integral to solve! Let's get this party started.

To find the integral of 3cos(2x) - 2sin(x), we can break it down into individual integrals. The integral of 3cos(2x) can be found by using the power rule for integration:

∫3cos(2x) dx = (3/2) ∫cos(2x) d(2x)

Using a handy-dandy substitution, let's replace 2x with u:

u = 2x
du = 2 dx
dx = du/2

Now, we can rewrite the integral as:

(3/2) ∫cos(u) (du/2)

Simplifying a bit further, we have:

(3/4) ∫cos(u) du

The integral of cos(u) is simply sin(u), so let's substitute that back in:

(3/4)sin(u)

Now, we need to go back to our original variable, x:

(3/4)sin(2x)

Now, let's move on to the second term, -2sin(x):

The integral of -2sin(x) is straightforward:

∫-2sin(x) dx = -2∫sin(x) dx

Using another substitution, let's replace x with v:

v = x
dv = dx

The integral becomes:

-2∫sin(v) dv

And the integral of sin(v) is just -cos(v), so let's plug that back in:

-2(-cos(v)) = 2cos(v)

Finally, let's replace v with x:

2cos(x)

Now, we can add it all up:

∫(3cos(2x) - 2sin(x)) dx = (3/4)sin(2x) + 2cos(x)

And there you have it! The integral of 3cos(2x) - 2sin(x) is (3/4)sin(2x) + 2cos(x). Enjoy the math party!

## To find the integral of 3cos(2x) - 2sin(x), you can use integration techniques and formulas.

Step 1: Observe the function and its constituents. In this case, we have two terms: 3cos(2x) and 2sin(x). Each term can be integrated separately.

Step 2: Integrate the first term, 3cos(2x). To integrate cos(2x), we can make use of the formula ∫cos(ax) dx = (1/a)sin(ax) + C, where C is the constant of integration. Applying this formula, we have:

∫3cos(2x) dx = (3/2)∫cos(2x) dx = (3/2)(1/2)sin(2x) + C1 = (3/4)sin(2x) + C1

Step 3: Integrate the second term, -2sin(x). To integrate sin(x), we can use the formula ∫sin(ax) dx = -(1/a)cos(ax) + C. Applying this formula, we have:

∫-2sin(x) dx = -2∫sin(x) dx = -2(-1)cos(x) + C2 = 2cos(x) + C2

Step 4: Combine the integrated terms. The integral of the original function is the sum of the integrals of each term:

∫(3cos(2x) - 2sin(x)) dx = (3/4)sin(2x) + 2cos(x) + C,

where C = C1 + C2 is the constant of integration.

Therefore, the integral of 3cos(2x) - 2sin(x) is (3/4)sin(2x) + 2cos(x) + C.

## Recall that cos (2x) = 2 cos^2 x - 1

or
cos^2 x = (cos 2x + 1)/2 = (1/2)cos 2x + 1/2

∫ 3cos2x-2sinx dx
= ∫ (3/2)cos 2x + 3/2 - 2sinx ) dx
= 3/4 sin 2x + (3/2)x + 2cosx + c