# Integration of sin square x: Formula, Proof | sin^2x Integration

The integration of sin square x is equal to x/2 – (sin2x)/4 +C. Here we learn how to integrate sin2x. The integration of sin2x formula is given by

∫sin2x dx = x/2 – (sin2x)/4 +C

where C is an arbitrary integral constant.

## How to Integrate sin^2x

To integrate sin square x, we will use the trigonometric formula of sin2x. We know that

cos2x = 1- 2sin2x.

This formula can be rewritten as follows:

2sin2x = 1 – cos2x

∴ sin2x = $\dfrac{1}{2}$(1 – cos 2x)

Integrating both sides, we get that

∫sin2x dx = $\dfrac{1}{2}$ ∫(1-cos 2x) dx + C where C is a constant of integration.

= $\dfrac{1}{2}$ ∫dx – $\dfrac{1}{2}$ ∫ cos 2x dx + C

[Let 2x=t. So we have 2dx = dt ⇒ dx =dt/2]

= $\dfrac{1}{2}$ x – $\dfrac{1}{2} \times \dfrac{1}{2}$ ∫ cos t dt + C

= $\dfrac{1}{2}$ x – $\dfrac{1}{4}$ sin t + C

= $\dfrac{x}{2}$ – $\dfrac{\sin 2x}{4}$ + C as t=2x.

So the integration of sin^2x (sine square x) is equal to x/2 – (sin2x)/4 +C where C is an arbitrary constant.

Solution:

From above, the integral of sin2x is equal to x/2 – (sin2x)/4 +C. Thus,

∫$_0^{\pi/2}$ sin2x dx = [x/2 – (sin2x)/4 +C]$_0^{\pi/2}$

= [π/4 – (sin π)/4 +C] – [0/2 – (sin0)/4+C]

= (π/4 – 0 +C) – (0 -0+C) as we know that sin π = 0 and sin 0 =0.

= π/4.

So the value of the integral of sin2x from 0 to π/2 is equal to π/4.