# Find the integral of 3cos2x-2sinx

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

in the required format on this webpage.

Bosnian interpreted your 3cos2x

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

I interpreted your 3cos2x

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

Make sure that you make it clear which way.

## If your question means:

∫ [ 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

## Tnx

## 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