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Integrate x·(pi - x)·sin(x) from 0 to pi
Step by step
- \int_{0}^{\pi} x \left(\pi - x\right) \sin{\left(x \right)}\, dx
First find an antiderivative F, then evaluate F(b) − F(a).
- u = - x,\quad du = -1\, dx
Substitute u = - x.
- \int x \left(\pi - x\right) \sin{\left(x \right)}\, dx = \int - u^{2} \sin{\left(u \right)} - u \pi \sin{\left(u \right)}\, d_u
Rewrite the integral in terms of u.
- \int - u^{2} \sin{\left(u \right)} - u \pi \sin{\left(u \right)}\, d_u = \int - u^{2} \sin{\left(u \right)}\, d_u + \int - u \pi \sin{\left(u \right)}\, d_u
The integral of a sum is the sum of the integrals.
- \int - u^{2} \sin{\left(u \right)}\, d_u = -1 \int u^{2} \sin{\left(u \right)}\, d_u
Pull the constant -1 out of the integral.
- u = u^{2},\quad dv = \sin{\left(u \right)}\, d_u
Integration by parts: ∫u dv = uv − ∫v du.
- du = 2 u\, d_u,\quad v = - \cos{\left(u \right)}
Differentiate u, integrate dv.
- \int u^{2} \sin{\left(u \right)}\, d_u = - u^{2} \cos{\left(u \right)} - \int - 2 u \cos{\left(u \right)}\, d_u
Apply the formula.
- u = - 2 u,\quad dv = \cos{\left(u \right)}\, d_u
Integration by parts: ∫u dv = uv − ∫v du.
- du = -2\, d_u,\quad v = \sin{\left(u \right)}
Differentiate u, integrate dv.
- \int - 2 u \cos{\left(u \right)}\, d_u = - 2 u \sin{\left(u \right)} - \int - 2 \sin{\left(u \right)}\, d_u
Apply the formula.
- \int - 2 \sin{\left(u \right)}\, d_u = -2 \int \sin{\left(u \right)}\, d_u
Pull the constant -2 out of the integral.
- \int \sin{\left(u \right)}\, d_u = - \cos{\left(u \right)}
Standard trigonometric antiderivative.
- = - 2 u \sin{\left(u \right)} - 2 \cos{\left(u \right)}
Combine.
- = - u^{2} \cos{\left(u \right)} + 2 u \sin{\left(u \right)} + 2 \cos{\left(u \right)}
Combine.
- \int - u \pi \sin{\left(u \right)}\, d_u = - \pi \int u \sin{\left(u \right)}\, d_u
Pull the constant - \pi out of the integral.
- u = u,\quad dv = \sin{\left(u \right)}\, d_u
Integration by parts: ∫u dv = uv − ∫v du.
- \vdots
…
- = x^{2} \cos{\left(x \right)} - 2 x \sin{\left(x \right)} - \pi \left(x \cos{\left(x \right)} - \sin{\left(x \right)}\right) - 2 \cos{\left(x \right)}
Substitute back u = - x.
- F(\pi) - F(0) = \left(2\right) - \left(-2\right)
Fundamental theorem of calculus: plug in the limits.
- = 4
Simplify.