Solve the integral from [0,pi] of x*sinx*cosx*dx
I don't understand how to solve this problem because if I let u =x and du =1 and dv=sinx*cosx, I can't find v.
How do you solve this?
x sinx cosx = 1/2 x sin2x
u = x/2
du = 1/2 dx
dv = sin2x dx
v = -1/2 cos2x
∫u dv = uv - v du
= -1/4 x cos2x + 1/4 ∫cos2x dx
= -1/4 x cos2x + 1/8 sin2x + C
using the limits of integration, we have
(-1/4 π + 0) - (0 +0)
= -π/4
I appreciate your kindness! Calc II is really a hard subject for me.
To solve the integral ∫[0, π] x*sin(x)*cos(x) dx, you can use integration by parts. Integration by parts is a technique that involves splitting the integral into two parts and creating a new integral involving the derivative of one function and the antiderivative of another function.
Let's first apply the integration by parts formula, which states:
∫ [a, b] u(x) v'(x) dx = [u(x)v(x)] [a, b] - ∫ [a, b] u'(x) v(x) dx,
where u(x) and v(x) are functions of x.
In this case, we can choose u(x) = x and dv(x) = sin(x)*cos(x) dx. To find v(x), we integrate dv(x). However, as you correctly pointed out, it is challenging to find the antiderivative of sin(x)*cos(x).
To overcome this, we can use the trigonometric identity sin(2x) = 2sin(x)cos(x). Rearranging, we have sin(x)*cos(x) = sin(2x)/2. Therefore, we substitute this expression for dv(x) in the integration by parts formula:
∫ [0, π] x*sin(x)*cos(x) dx = ∫ [0, π] x * (sin(2x)/2) dx.
Now, we can find the integral of the new expression:
∫ [0, π] x * (sin(2x)/2) dx.
To solve this, you can use integration by parts again. Choose u(x) = x and dv(x) = (sin(2x)/2) dx. Differentiate u(x) to find u'(x), and integrate dv(x) to find v(x).
After finding u'(x) and v(x), apply the integration by parts formula:
∫ [0, π] x * (sin(2x)/2) dx = [x * v(x)] [0, π] - ∫ [0, π] v(x) dx.
Now, you have a new integral involving v(x). Evaluate the limits of integration, and subtract the integral of v(x) from the first term on the right side of the equation.
This new integral may require further simplification or the use of additional trigonometric identities. Repeat the process of integration by parts or apply other methods as needed until you can evaluate the integral completely.
Remember to be patient and persistent, as integration problems can sometimes involve multiple steps and require the use of various techniques.