That's the same as the integral of sin^2 x dx.
Use integration by parts.
Let sin x = u and sin x dx = dv
v = -cos x
du = cos x dx
The integral is u v - integral of v du
= -sinx cosx + integral of cos^2 dx
which can be rewritten
integral of sin^2 x = -sinx cos x + integral of (1 - sin^2) dx
2 * (integral of sin^2 x dx)
= - sin x cos x + integral of dx
integral of sin^2 dx = (-1/2) sin x cos x + x/2
integral of (1-cos^2 x) dx
might be easy but i need to make sure
To find the integral of (1 - cos^2 x) dx, you can simplify the expression first.
Using the identity for cosine squared (cos^2 x = 1 - sin^2 x), we can rewrite the expression as:
(1 - cos^2 x) dx = sin^2 x dx
Now, we can go ahead and integrate sin^2 x dx using the method of integration by parts, as mentioned earlier.
Let's repeat the steps:
1. Start by letting u = sin x and dv = sin x dx.
This gives us du = cos x dx and v = -cos x.
2. Apply the integration by parts formula:
∫ u dv = uv - ∫ v du
Plugging in the values, we have:
∫ sin^2 x dx = -sin x cos x - ∫ (-cos x)(cos x dx)
3. Simplify the expression:
∫ sin^2 x dx = -sin x cos x + ∫ cos^2 x dx
4. Now we have another integral to solve. Use the identity for cosine squared (cos^2 x = 1 - sin^2 x) to rewrite the expression:
∫ sin^2 x dx = -sin x cos x + ∫ (1 - sin^2 x) dx
Distribute the integral over the addition:
∫ sin^2 x dx = -sin x cos x + ∫ dx - ∫ sin^2 x dx
The integral of dx is x:
∫ sin^2 x dx = -sin x cos x + x - ∫ sin^2 x dx
5. Rearrange the equation to isolate the integral of sin^2 x dx:
2 ∫ sin^2 x dx = -sin x cos x + x
6. Divide both sides by 2:
∫ sin^2 x dx = (-1/2) sin x cos x + x/2
So the final result is (-1/2) sin x cos x + x/2.
Remember to always double-check your answer by differentiating the result to see if you get the original function.