[Solved] What is the Antiderivative of sin2x?

The antiderivative of sin2x is (-cos2x)/2+C where ln denotes the logarithm with base e and C is a constant. In this post, we will learn what is the antiderivative of sin2x.

The antiderivative formula of sin2x is given by

∫sin2x dx = $-\dfrac{\cos 2x}{2}$ + C,

where C is an integration constant.

How to Find the Antiderivative of sin2x?

The antiderivative of sin2x is a function whose derivative is sin2x. Thus, the antiderivative of sin2x = the integration of sin2x.

Now,

∫ sin2x dx.

Let 2x = z

Differentiating both sides with respect to x, we get

2 = dz/dx

⇒ dx = dz/2

Thus the integral of sin2x is

∫ sin2x dx

= ∫ sinz dz/2

= 1/2 ∫ sinz dz + C

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

= (-cos2x)/2+C as z=2x.

Hence, ∫sin2x dx = (-cos2x)/2+C.

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

Read the Antiderivatives of:

1	1/x	x
ln x	tan x	cot x
sin2x	cos2x	lnx/x

FAQs