The antiderivative of cos2x is (sin2x)/2+C where ln denotes the logarithm with base e and C is a constant. Let us learn how to find the antiderivative of cos2x.

The antiderivative formula of cos2x is given by

∫cos2x dx = $\dfrac{\sin 2x}{2}$ + C,

where C is an integration constant.

## Find Antiderivative of cos2x

The antiderivative of cos2x is a function whose derivative is cos2x. Thus, the antiderivative of cos2x equals the integration of cos2x.

Now,

∫ cos2x dx.

Let 2x = t

Differentiating both sides with respect to x, we get

2 = dt/dx

⇒ dx = dt/2

Thus the integral of cos2x is

∫cos2x dx

= ∫cost dt/2

= 1/2 ∫cost dt + C

= (sint)/2 + C as the integral of cosx is sinx.

= (sin2x)/2+C as t=2x.

Hence, ∫cos2x dx = (sin2x)/2+C.

So the antiderivative of cos2x is (sin2x)/2+C where C is a constant on integration.

**Verification:**

Let’s verify that the derivative of (sin2x)/2+C equals cos2x. Now, the derivative of (sin2x)/2+C is equal to

$\frac{d}{dx}$[(sin2x)/2+C]

= $\frac{1}{2} \frac{d}{dx}$(sin2x+C)

= 1/2 ⋅ 2 cos2x + 0 as the derivative of a constant is zero and the derivative of sinx is cosx.

= cos2x, hence verified.

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## FAQs

### Q1: What is antiderivative of cos2x?

Answer: The antiderivative of cos2x is (sin2x)/2+C where C is an integration constant.