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.