What is the integral of sin^4 (x) dx?

Assume that, I = ∫ sin⁴x dx

Now, reduce sin4x dx to integrate the function.

I = ∫ (sin²x)² dx

As we know 2sin²x = 1 – cos2x,

Put (sin²x)² = {(1 − cos2x) / 2}²

= ∫ {(1 − cos2x) /2 }² dx

= (1/4) ∫ (1 − cos2x)² dx

As, (a – b)² = a² – 2ab + b²

= (1/4) ∫ (1− 2cos2x + cos²2x) dx .(1)

Again reduce the function cos²2x to integrate the function,

As, cos²(2x) = (1 + cos(4x)) / 2 … (2)

Now, substitute the value (2) in (1), we get

= (1/4) ∫ [1 – 2cos2x + {(1 + cos(4x)) / 2 }] dx

Now, integrate the function, we get

= (1/32)(12x – 8 sin (2x) + sin (4x) )+C

On simplifying the above expression, we get

= (3/8)x + (1/32) sin(4x) − (1/4) sin(2x) + C

Therefore, the integral of sin4x dx = (3/8)x + (1/32) sin(4x) − (1/4) sin(2x) + C