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.