Integration of Sin2x

Integration of sin2x means finding the integral of the function sin2x. Integral of sin2x can be written as ∫ sin2x dx. Here, we need to find the indefinite integral of sin2x. So, the integration of sin2x results in a new function with arbitrary constant C. Let’s learn the formula for the integration of sin2x along with the derivation.

Integration of Sin2x dx Formula

The formula for the integral of sin 2x dx is given by:

Learn: Integral

Integration of Sin 2x Derivation

We can derive the formula of integral of sin2x using the method of integration by substitution.

Consider ∫sin2x dx

Let u = 2x such that du = 2dx ⇒ dx = (½)du.

Substituting the above equations in ∫sin2x dx, we get;

∫ sin2x dx = ∫ sin u (½) du

= (½) ∫ sin u du

= (½) (-cos u) + C

= -(½) cos2x + C

Therefore, ∫ sin2x dx = -(½) cos2x + C

Solved Examples

Let’s have a look at the examples given below to understand how to apply the formula of integration of sin 2x.

Example 1: Integration of Sin2x dx from 0 to pi/2

Integration of sin2x dx from 0 to pi/2 can be written as:

This is the definite integral of sin2x with lower limit 0 and upper limit π/2.

We know that, ∫ sin2x dx = -(½) cos2x + C

So,

Now, apply the limits to the new function.

= -(½) [cos π – cos 0]

-(½) [-1 – 1]

= -(½)(-2)

= 1

Therefore, integration of sin 2x from o to pi/2 is equal to 1.

Example 2: Integration of Sin(2x+1)

Integration of sin(2x+1) can be written as: ∫ sin(2x + 1)dx

We know that, ∫ sin2x dx = -(½) cos2x + C

So, ∫ sin(2x + 1) dx = -(½) cos(2x+1) + C

Example 3: Integration of Sin2x/1+cosx

Integration of Sin2x/1+cosx

= ∫ (sin2x)/(1 + cos x) dx

Using the sin2x formula, i.e. sin2x = 2 sinx cosx

= ∫ (2 sinx cosx)/(1 + cosx) dx

= 2 ∫[cosx/(1 + cos x)] sinx dx

Let u = cos x

du = -sinx dx

Susbtituting these values, we get;

= 2 ∫[u/(1 + u)] (-du)

= -2 ∫ (u + 1 – 1)/(u + 1) du

= -2 ∫ (u + 1)/(u + 1) du + 2 ∫ 1/(u + 1) du

= -2 ∫ 1 du + 2 log|u + 1|

= -2u + 2log|u + 1| + C

= -2 cosx + 2log|cosx + 1| + C