How do you integrate #(cosx)^2dx#?

Answer 1

#int cos^2xdx = 1/2(x+1/2sin2x) + C#

Considering that:

#(dsinx)/dx = cosx#
#(dcosx)/dx = -sinx#
#int cos^2xdx =int cosx * cosx dx =int cosx d(sinx)#

Integrating by parts:

#int cos^2xdx = sinxcosx - int sinx dcosx = #
# = sinxcosx + int sin^2x dx = #
# = sinxcosx + int (1-cos^2x) dx = #
# = sinxcosx + x - int cos^2x dx #

So:

#2int cos^2xdx = sinxcosx + x#

and finally:

#int cos^2xdx = 1/2(x+1/2sin2x) + C#

An alternative method is to use the identity:

#cos(2x) = cos^2x-sin^2x = cos^2x - (1-cos^2x) = # # = 2cos^2x-1#

so that:

#cos^2x = (1+cos2x)/2#
#int cos^2xdx = int (1+cos2x)/2 dx = 1/2(x+1/2sin2x) + C#
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Answer 2

To integrate (cosx)^2 dx, you can use the trigonometric identity:

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

Then, the integral becomes:

∫ (cosx)^2 dx = ∫ (1 + cos(2x))/2 dx.

This simplifies to:

(1/2) ∫ (1 + cos(2x)) dx.

Integrate term by term:

(1/2) ∫ dx + (1/2) ∫ cos(2x) dx.

Integrating each term:

(1/2) * x + (1/4) * sin(2x) + C.

So, the integral of (cosx)^2 dx is:

(1/2) * x + (1/4) * sin(2x) + C, where C is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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