Question

Especially for large and even powers, integrating sine powers can be difficult.

(a) Determine $\int {\sin }^{4}xdx,\int {\sin }^{8}xdx$ using the provided reduction formula. The formula might need to be used more than once.

(b) What does the term "reduction formula" mean?

(c) Show the reduction formula using integration by parts.

Short answer

1. $$\begin{gathered} \int {\sin }^{4}xdx=-\frac{1}{4}{\sin }^{3}x\cos x+\frac{3}{8}x-\frac{3}{16}\sin 2x+C \\ \int {\sin }^{8}xdx=-\frac{1}{8}{\sin }^{7}x\cos x-\frac{7}{48}{\sin }^{5}x\cos x-\frac{35}{192}{\sin }^{3}x\cos x+\frac{35}{128}x-\frac{35}{256}\sin 2x+C \end{gathered}$$
2. Reduction formula reduces the power of $\sin x.$
3. Applying integral by parts formula we get $(k-1)\int {\sin }^{k}xdx$

Step-by-step solution

Part(a) Step 1: Given information

The integral $\int {\sin }^{4}xdx,\int {\sin }^{8}xdx$

Part(a) Step 2: Calculation

The objective is to solve the integration

$\int {\sin }^{4}xdx$

$=-\frac{1}{4}{\sin }^{3}x\cos x+\frac{4-1}{4}\int {\sin }^{2}xdx=-\frac{1}{4}{\sin }^{3}x\cos x+\frac{3}{4}\int {\sin }^{2}xdx=-\frac{1}{4}{\sin }^{3}x\cos x+\frac{3}{8}x-\frac{3}{16}\sin 2x+C$

Therefore, the value is $-\frac{1}{4}{\sin }^{3}x\cos x+\frac{3}{8}x-\frac{3}{16}\sin 2x+C$

$\int {\sin }^{8}xdx$

$$\begin{gathered} =-\frac{1}{8}{\sin }^{7}x\cos x+\frac{8-1}{8}\int {\sin }^{6}xdx \\ =-\frac{1}{8}{\sin }^{7}x\cos x+\frac{7}{8}\int {\sin }^{6}xdx \\ =-\frac{1}{8}{\sin }^{7}x\cos x-\frac{7}{48}{\sin }^{5}x\cos x \\ -\frac{35}{192}{\sin }^{3}x\cos x+\frac{35}{128}x-\frac{35}{256}\sin 2x+C \end{gathered}$$

Therefore, the value is

$$\begin{gathered} -\frac{1}{8}{\sin }^{7}x\cos x-\frac{7}{48}{\sin }^{5}x\cos x \\ -\frac{35}{192}{\sin }^{3}x\cos x+\frac{35}{128}x-\frac{35}{256}\sin 2x+C \end{gathered}$$

Part(b) Step 1: Given information

The function $\sin x$

Part(b) Step 2: Explanation

Every time we use the formula, the power of $\sin x$ in the integrand is reduced by two.

Part(c) Step 1: Given information

To prove the reduction formula

Part(c) Step 2: Calculation

To demonstrate the reduction formula, use integration by parts.

$$\begin{gathered} u={\sin }^{k-1}x \\ du=(k-1){\sin }^{k-2}x\cos xdx \\ dv=\sin xdx \\ v=-\cos x \end{gathered}$$

Multiply the integrand after using the integration by parts formula to get a term of the form,

$(k-1)\int {\sin }^{k}xdx$

And then solve for$\int {\sin }^{k}xdx.$