The solution to the integral of sin^2(x) requires you to recall principles of both trigonometry and calculus. Don't conclude that since the integral of sin(x) equals -cos(x), the integral of sin^2(x) should equal -cos^2(x); in fact, the answer does not contain a cosine at all. You cannot directly integrate sin^2(x). Use trigonometric identities and calculus substitution rules to solve the problem.
Int sin^2xcosxdx = 1/3sin^3x + C Let I = int sin^2xcosxdx We can integrate this by substitution: Let u=sinx = (du)/dx=cosx = int. Cosxdx = int. Du And so, we can rewrite the integral as follows: I = int (sin^2x)cosxdx I = int u^2du I = 1/3u^3 Ad re-substituting for u we gte: I = 1/3sin^3x + C. For example you can integrate Integral of 1,x^2 sin(t) dt as: sin(x^2) d/dx (x^2) = sin(x^2) (2x) = 2x sin(x^2) The lower limit of integration is 1 ( a constant). The upper limit of integration.
- For a definite integral, eliminate the constant in the answer and evaluate the answer over the interval specified in the problem. If the interval is 0 to 1, for example, evaluate [1/2 - sin(1)/4] - [0/2 - sin(0)/4)].
Use the half angle formula, sin^2(x) = 1/2*(1 - cos(2x)) and substitute into the integral so it becomes 1/2 times the integral of (1 - cos(2x)) dx.
Set u = 2x and du = 2dx to perform u substitution on the integral. Since dx = du/2, the result is 1/4 times the integral of (1 - cos(u)) du.
Integrate the equation. Since the integral of 1du is u, and the integral of cos(u) du is sin(u), the result is 1/4*(u - sin(u)) + c.
Substitute u back into the equation to get 1/4*(2x - sin(2x)) + c. Simplify to get x/2 - (sin(x))/4 + c.
![Sin^2x Sin^2x](/uploads/1/2/4/7/124796929/967915039.jpeg)
To integrate sin2x, also written as ∫sin2x dx, and sin 2x, we usually use a u substitution to build a new integration in terms of u.
Let u=2x.
Then du/dx = 2
We rearrange to get an expression for dx in terms of u.
As you can see, we now have a new integration in terms of u, which means the same thing. We get this by replacing 2x with u, and replacing dx with ½ du.
We move the ½ outside of the integral as it is simply a multiplier.
We now have a simple integration with sinu that we can solve easily.
We remember the ½ outside the integral sign and reintroduce it here.
Hence, this is the final answer -½cos2x + C, where C is the integration constant.