Question

integral of sin^3 x / 1+cos x

Answer 1

Answer on Question #44093 – Math – Integral Calculus

Integral of $\sin^3 x / (1 + \cos x)$

Solution

$$\int \frac{\sin^3 x \, dx}{1 + \cos(x)} = \int \frac{\sin^2 x \cdot \sin(x) \, dx}{1 + \cos(x)} = -\int \frac{\sin^2 x \cdot (-\sin(x)) \, dx}{1 + \cos(x)} = \left| \text{use} \right. \, d(\cos(x)) = -\sin(x) \, dx = \\ = -\int \frac{\sin^2 x \, d(\cos(x))}{1 + \cos(x)} = \\ = -\int \frac{(1 - \cos^2 x) \, d(\cos(x))}{1 + \cos(x)} = \left| \text{use} \right. \, a^2 - b^2 = (a - b)(a + b) \\ = -\int \frac{(1 - \cos(x))(1 + \cos(x)) \, d(\cos(x))}{1 + \cos(x)} = \left| \text{use} \right. \, \int -f(x) \, dx = -\int f(x) \, dx \\ = \int \left( -(1 - \cos(x)) \right) \, d(\cos(x)) = \\ = \left| \text{use} \right. \, d(-\cos(x)) = -d(\cos(x)) = \int (1 - \cos(x)) \, d(-\cos(x)) = \left| \text{use} \right. \, d(f(x) + g(x)) = \\ d(f(x)) + d(g(x)) = f'(x) \, dx + g'(x) \, dx, \quad (1)' = 0 = \int (1 - \cos(x)) \, d(1 - \cos(x)) = \\ \left| \text{substitution} \right. \, t = 1 - \cos(x) = \int t \, dt = \frac{t^2}{2} + C = \frac{(1 - \cos(x))^2}{2} + C = \frac{(2 \sin^2 \left(\frac{x}{2}\right))^2}{2} + C = \frac{4 \sin^4 \left(\frac{x}{2}\right)}{2} + C = \\ = 2 \sin^4 \left(\frac{x}{2}\right) + C,$$

where $C$ is an arbitrary real constant. Moreover, $\cos(x) \neq -1$, $x \neq \pi + 2k\pi$, $k$ is integer.

Answer: $2 \sin^4 \left(\frac{x}{2}\right) + C$

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