We can use the identity sin² x = 1/2(1-cos(2x)) to integrate sin² x dx as follows:
∫sin² x dx = ∫1/2(1-cos(2x)) dx
= 1/2(x - 1/2 sin(2x)) + C, where C is the constant of integration.
Integrate sin² x dx
3 answers
Integrate sin⁴x dx
We can use the identity sin²x = (1/2)(1-cos(2x)) to express sin⁴x in terms of cos(2x) as follows:
sin⁴x = (sin²x)²
= [(1/2)(1-cos(2x))]²
= (1/4)(1-2cos(2x)+cos²(2x))
Now, we can integrate sin⁴x dx using this expression:
∫sin⁴x dx = ∫(1/4)(1-2cos(2x)+cos²(2x)) dx
= (1/4)(x - sin(2x) + (1/2)(1/2)(2x + sin(4x))) + C
= (1/4)(x - sin(2x) + x/2 + 1/4sin(4x)) + C
= (3/8)x - (1/4)sin(2x) + (1/16)sin(4x) + C
Therefore, the integral of sin⁴x dx is (3/8)x - (1/4)sin(2x) + (1/16)sin(4x) + C.
sin⁴x = (sin²x)²
= [(1/2)(1-cos(2x))]²
= (1/4)(1-2cos(2x)+cos²(2x))
Now, we can integrate sin⁴x dx using this expression:
∫sin⁴x dx = ∫(1/4)(1-2cos(2x)+cos²(2x)) dx
= (1/4)(x - sin(2x) + (1/2)(1/2)(2x + sin(4x))) + C
= (1/4)(x - sin(2x) + x/2 + 1/4sin(4x)) + C
= (3/8)x - (1/4)sin(2x) + (1/16)sin(4x) + C
Therefore, the integral of sin⁴x dx is (3/8)x - (1/4)sin(2x) + (1/16)sin(4x) + C.