Question

Find the reduction formula of ʃsin10xdx.

Answer 1

Answer on Question #53130 – Math – Integral Calculus

Find the reduction formula of $\int \sin 10x \, dx$.

Solution

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

Thus

$$\begin{aligned} I_{10} = -\frac{1}{10} \sin^9 x \cos x + \frac{9}{10} I_8 = \\ = -\frac{1}{10} \sin^9 x \cos x + \frac{9}{10} \left(-\frac{1}{8} \sin^7 x \cos x + \frac{7}{8} I_6\right) = \\ = -\frac{1}{10} \sin^9 x \cos x - \frac{9}{80} \sin^7 x \cos x + \frac{63}{80} \left(-\frac{1}{6} \sin^5 x \cos x + \frac{5}{6} I_4\right) = \\ = -\frac{1}{10} \sin^9 x \cos x - \frac{9}{80} \sin^7 x \cos x - \frac{63}{480} \sin^5 x \cos x + \frac{315}{480} \left(-\frac{1}{4} \sin^3 x \cos x + \frac{3}{4} I_2\right) = \\ = -\frac{1}{10} \sin^9 x \cos x - \frac{9}{80} \sin^7 x \cos x - \frac{63}{480} \sin^5 x \cos x - \frac{315}{1920} \sin^3 x \cos x + \frac{945}{1920} \left(-\frac{1}{2} \sin x \cos x + \frac{1}{2} I_0\right) = \\ = -\frac{1}{10} \sin^9 x \cos x - \frac{9}{80} \sin^7 x \cos x - \frac{21}{160} \sin^5 x \cos x - \frac{105}{640} \sin^3 x \cos x - \frac{315}{1280} \sin x \cos x + \frac{315}{1280} I_0 \end{aligned}$$

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

So

$$\int \sin^{10}x \, dx =$$ $$= -\frac{\sin 2x}{2560} \left(256 \sin^{8}x + 144 \sin^{6}x + 168 \sin^{4}x + 210 \sin^{2}x + 315\right) + \frac{315}{2560} x + c$$

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