Question

Find the reduction formula of ʃsin8xdx.

Answer 1

Answer on Question #53128 – Math – Integral Calculus

Find the reduction formula of $\int \sin 8x \, dx$.

Solution

$$\begin{array}{l} I_n = \int \sin^n x \, dx = \int \sin^{n-1} x \sin x \, dx = - \int \sin^{n-1} x \, d(\cos x) = \\ = - \sin^{n-1} x \cos x + (n - 1) \int \sin^{n-2} x \cos^2 x \, dx = \\ = - \sin^{n-1} x \cos x + (n - 1) \int \sin^{n-2} x (1 - \sin^2 x) \, dx = \\ = - \sin^{n-2} x \cos x + (n - 1) I_{n-2} - (n - 1) I_n \rightarrow \\ \rightarrow I_n = - \frac{1}{n} \sin^{n-1} x \cos x + \frac{n - 1}{n} I_{n-2} \end{array}$$

Thus

$$\begin{array}{l} I_8 = - \frac{1}{8} \sin^7 x \cos x + \frac{7}{8} I_6 = - \frac{1}{8} \sin^7 x \cos x + \frac{7}{8} \left(- \frac{1}{6} \sin^5 x \cos x + \frac{5}{6} I_4\right) = \\ = - \frac{1}{8} \sin^7 x \cos x - \frac{7}{48} \sin^5 x \cos x + \frac{35}{48} \left(- \frac{1}{4} \sin^3 x \cos x + \frac{3}{4} I_2\right) = \\ = - \frac{1}{8} \sin^7 x \cos x - \frac{7}{48} \sin^5 x \cos x - \frac{35}{192} \sin^3 x \cos x + \frac{105}{192} \left(- \frac{1}{2} \sin x \cos x + \frac{1}{2} I_0\right) = \\ = - \frac{1}{8} \sin^7 x \cos x - \frac{7}{48} \sin^5 x \cos x - \frac{35}{192} \sin^3 x \cos x - \frac{105}{384} \sin x \cos x + \frac{105}{384} I_0 \end{array}$$

But $I_0 = \int dx = x + c$, where $c$ is an arbitrary real constant.

So

$$\int \sin^8 x \, dx = - \frac{\sin x \cos x}{384} \left(48 \sin^6 x + 56 \sin^4 x + 70 \sin^2 x + 105\right) + \frac{105}{384} x + c$$

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