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Integrate -4·sin(t)^2·cos(t) + cos(t) from 0 to 2·pi
\int_{0}^{2 \pi} - 4 \sin^{2}{\left(t \right)} \cos{\left(t \right)} + \cos{\left(t \right)}\, dt
Step by step
- \int_{0}^{2 \pi} - 4 \sin^{2}{\left(t \right)} \cos{\left(t \right)} + \cos{\left(t \right)}\, dt
First find an antiderivative F, then evaluate F(b) − F(a).
- \int - 4 \sin^{2}{\left(t \right)} \cos{\left(t \right)} + \cos{\left(t \right)}\, dt = \int - 4 \sin^{2}{\left(t \right)} \cos{\left(t \right)}\, dt + \int \cos{\left(t \right)}\, dt
The integral of a sum is the sum of the integrals.
- \int - 4 \sin^{2}{\left(t \right)} \cos{\left(t \right)}\, dt = -4 \int \sin^{2}{\left(t \right)} \cos{\left(t \right)}\, dt
Pull the constant -4 out of the integral.
- u = \sin{\left(t \right)},\quad du = \cos{\left(t \right)}\, dt
Substitute u = \sin{\left(t \right)}.
- \int \sin^{2}{\left(t \right)} \cos{\left(t \right)}\, dt = \int u^{2}\, d_u
Rewrite the integral in terms of u.
- \int u^{2}\, d_u = \frac{u^{3}}{3}
Power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) (n ≠ −1).
- = \frac{\sin^{3}{\left(t \right)}}{3}
Substitute back u = \sin{\left(t \right)}.
- \int \cos{\left(t \right)}\, dt = \sin{\left(t \right)}
Standard trigonometric antiderivative.
- F(2 \pi) - F(0) = \left(0\right) - \left(0\right)
Fundamental theorem of calculus: plug in the limits.
- = 0
Simplify.
Reveal the answer
0