can someone show me step by step how to evaluate the integral of 2(sinx)^2 dx

I just don't understand how to do it....

Isn't there are identity

sin2x=1-2sin^2 x ?

I don't understand what you are saying. there is no sin2x in my equation i don't think

i have to evaluate the integral 2(sin^2x)

Try

sin(x)sin(y)=(1/2)(cos(x-y)-cos(x+y))
Put y=x and noting that cos(0)=1, you will have a simpler expression to integrate.

Sure! I can help you with that. To evaluate the integral of 2(sin x)^2 dx, we can use a combination of trigonometric identities and integration techniques.

Step 1: Start by using the double angle identity for sine: sin^2 x = (1 - cos 2x) / 2.

Step 2: Rewrite the integral as 2 * (1 - cos 2x) / 2 dx.

Step 3: Simplify the expression: (1 - cos 2x) dx.

Step 4: Distribute the 2 into the expression to obtain 2 dx - 2cos 2x dx.

Step 5: Separate the integral into two parts: ∫2 dx - ∫2cos 2x dx.

Step 6: Evaluate the first integral: ∫2 dx = 2x + C, where C is the constant of integration.

Step 7: Evaluate the second integral: ∫-2cos 2x dx. This requires using the substitution method. Let u = 2x, then du/dx = 2, and dx = du/2.

Step 8: Substitute the values back into the integral: ∫-2cos 2x dx = ∫-2 cos u (du/2).

Step 9: Simplify the expression: -∫cos u du.

Step 10: Evaluate the integral of cos u: -sin u + C = -sin 2x + C.

Step 11: Combine the results of the two integrals: 2x - sin 2x + C.

So, the integral of 2(sin x)^2 dx is 2x - sin 2x + C, where C is the constant of integration.