Question

integral of sin^3 (2x) / 1+ cos(2x)

Answer 1

Answer on Question #44097 – Math – Integral Calculus

Integral of $\sin^3 (2x) / (1 + \cos (2x))$

Solution

$$\int \frac {\sin^ {3} (2 x) d x}{1 + \cos (2 x)} = \int \frac {\sin^ {2} (2 x) \cdot \sin (2 x) d x}{1 + \cos (2 x)} = - \frac {1}{2} \int \frac {\sin^ {2} (2 x) \cdot (- \sin (2 x)) d (2 x)}{1 + \cos (x)} =$$ $$| u s e d (2 x) = 2 d (x), \qquad d (\cos (2 x)) = - 2 \sin (2 x) d x | =$$ $$\begin{array}{l} = - \frac {1}{2} \int \frac {\sin^ {2} (2 x) d (\cos (2 x))}{1 + \cos (2 x)} = \\ = - \frac {1}{2} \int \frac {(1 - \cos^ {2} (2 x)) d (\cos (2 x))}{1 + \cos (2 x)} = | u s e a ^ {2} - b ^ {2} = (a - b) (a + b) | \\ = - \frac {1}{2} \int \frac {(1 - \cos (2 x)) (1 + \cos (2 x)) d (\cos (2 x))}{1 + \cos (2 x)} = | u s e \int - f (x) d x = - \int f (x) d x | \\ = \frac {1}{2} \int \left(- (1 - \cos (2 x))\right) d (\cos (2 x)) = \\ = | u s e d (- \cos (2 x)) = - d (\cos (2 x)) | = \frac {1}{2} \int (1 - \cos (2 x)) d (- \cos (2 x)) = \\ \end{array}$$ $$\begin{array}{l} \left| u s e d (f (x) + g (x)) = d (f (x)) + d (g (x)) = f ^ {\prime} (x) d x + g ^ {\prime} (x) d x, (1) ^ {\prime} = 0 \right| = \\ = \frac {1}{2} \int (1 - \cos (2 x)) d (1 - \cos (2 x)) = | s u b s t i t u t i o n t = 1 - \cos (2 x) | = \frac {1}{2} \int t d t = \frac {1}{2} \frac {t ^ {2}}{2} + C = \\ \frac {(1 - \cos (2 x)) ^ {2}}{4} + C = \frac {(2 \sin^ {2} (x)) ^ {2}}{4} + C = \frac {4 \sin^ {4} (x)}{4} + C = \sin^ {4} (x) + C, \\ \end{array}$$

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

$$x \neq \frac {\pi}{2} + k \pi , k \text{ is integer}.$$

Answer: $\sin^4 (x) + C$

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