Question

find antiderivative of sin2x​

Answer 1

Answer:

Antiderivative of sin 2x

Now, d/dx of cos 2x = (-sin 2x).2 = -2 sin 2x

So, d/dx of -(1/2)cos 2x = -(1/2).(-2 sin 2x) = sin 2x.

Thus the antderivative of sin 2x is -(1/2)cos 2x.

OR

Integral [sin 2x] dx = ?

Let t = 2x

dt/dx = 2

dx = dt/2

Substituting t,we get -

Integral[sin 2x] dx = Integral[sin t] dt/2

= (1/2)Integral[sin t] dt = (1/2)[ -cos t + k]

= (-1/2)cos t + C (C = 1/2 k)

= (-1/2)cos 2x + C.

Answer 2

${\large{\bold{\sf{➠\:∫sin2x}}}}$

${\large{\bold{\sf{➠\:-\frac{1}{2}cos2x+c }}}}$