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Found in: Page 340

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Evaluate the integral $$\int {{{{\mathop{\rm Sin}\nolimits} }^6}2x.dx}$$

$$\frac{1}{{384}}(120x - 33{\mathop{\rm Sin}\nolimits} 4x + 9{\mathop{\rm Sin}\nolimits} 8x - {\mathop{\rm Sin}\nolimits} 12x)$$+c is the value of given integral

See the step by step solution

## Step 1- Use the Formula \begin{aligned}{l}{\mathop{\rm Cos}\nolimits} 2y = 1 - 2{{\mathop{\rm Sin}\nolimits} ^2}y\\{\mathop{\rm Cos}\nolimits} 2y = 2{{\mathop{\rm Cos}\nolimits} ^2}y - 1\end{aligned}

$$\int {{{{\mathop{\rm Sin}\nolimits} }^6}2x.dx = \int {{{({{{\mathop{\rm Sin}\nolimits} }^2}2x)}^2}({{{\mathop{\rm Sin}\nolimits} }^2}2x).dx} }$$

=$$\int {(\frac{{1 - {\mathop{\rm Cos}\nolimits} (2(2x))}}{2}} {)^2}*(\frac{{1 - {\mathop{\rm Cos}\nolimits} 2(2x)}}{2}).dx$$

\begin{aligned}{l}{\int {\frac{1}{8}(1 - {\mathop{\rm Cos}\nolimits} 4x)} ^2}*(1 - {\mathop{\rm Cos}\nolimits} 4x).dx\\ = \frac{1}{8}\int {{{(1 - {\mathop{\rm Cos}\nolimits} 4x)}^3}} \\\end{aligned}

Because;$${(A - B)^3} = {A^3} - {B^3} - 3{A^2}B + 3A{B^2}$$

Therefore, The integral would become,

=$$\frac{1}{8}\int {(1 - {{{\mathop{\rm Cos}\nolimits} }^3}4x} - 3{\mathop{\rm Cos}\nolimits} 4x + 3{{\mathop{\rm Cos}\nolimits} ^2}4x).dx$$

=$$\frac{1}{8}\int {1 - (\frac{{{\mathop{\rm Cos}\nolimits} 8x + 1}}{2})*{\mathop{\rm Cos}\nolimits} 4x - 3{\mathop{\rm Cos}\nolimits} 4x + 3(\frac{{{\mathop{\rm Cos}\nolimits} 8x + 1}}{2}} ).dx$$

## Step 2- Evaluating of integral

=$$\frac{1}{8}(\int {dx - \int {(\frac{{{\mathop{\rm Cos}\nolimits} 8x*{\mathop{\rm Cos}\nolimits} 4x}}{2}} + \frac{{{\mathop{\rm Cos}\nolimits} 4x}}{2}} - 3{\mathop{\rm Cos}\nolimits} 4x).dx + \frac{3}{2}\int {({\mathop{\rm Cos}\nolimits} 8x + 1).dx}$$

=$$\frac{1}{8}(x - \int {\frac{1}{4}\{ {\mathop{\rm Cos}\nolimits} (8x + 4x) + {\mathop{\rm Cos}\nolimits} (8x - 4x)\} .dx} + \int {(\frac{1}{2} - 3){\mathop{\rm Cos}\nolimits} 4x.dx + \frac{3}{2}(\frac{{{\mathop{\rm Sin}\nolimits} 8x}}{8} + x)) + {c_1}}$$

= $$\frac{1}{8}(x - \frac{1}{4}(\int {{\mathop{\rm Cos}\nolimits} 12x + {\mathop{\rm Cos}\nolimits} 4x).dx - \frac{5}{2}\frac{{{\mathop{\rm Sin}\nolimits} 4x}}{4} + \frac{3}{{26}}{\mathop{\rm Sin}\nolimits} 8x + \frac{{3x}}{2})} + {c_2}$$

=$$\begin{array}{l}\frac{1}{8}(\frac{{( - )5x}}{2} - \frac{{{\mathop{\rm Sin}\nolimits} 12}}{{48}} - \frac{{11{\mathop{\rm Sin}\nolimits} 4x}}{{16}} + \frac{{3{\mathop{\rm Sin}\nolimits} 8x}}{{16}}) + c\\\end{array}$$

=$$\frac{1}{{384}}(120x - 33{\mathop{\rm Sin}\nolimits} 4x + 9{\mathop{\rm Sin}\nolimits} 8x - {\mathop{\rm Sin}\nolimits} 12x)$$..

Hence,$$\int {{{{\mathop{\rm Sin}\nolimits} }^6}2x.dx}$$= $$\frac{1}{{384}}(120x - 33{\mathop{\rm Sin}\nolimits} 4x + 9{\mathop{\rm Sin}\nolimits} 8x - {\mathop{\rm Sin}\nolimits} 12x)$$