What is the integral of $\int {sin}^{4} (2 x) d x$ $int sin^(4) (2x) dx$ ?

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What is the integral of $\int {sin}^{4} (2 x) d x$ $int sin^(4) (2x) dx$ ?

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Answer 1 · 2018-04-24T06:45:01

$\int {sin}^{4} (2 x) d x = \frac{3 x}{8} - \frac{sin (4 x)}{8} + \frac{sin (8 x)}{64} + C$ $int sin^4(2x) dx= (3x)/8-(sin(4x))/8 + (sin(8x))/64+C$

Explanation:

Use the trigonometric identities:

${sin}^{2} α = \frac{1 - cos 2 α}{2}$ $sin^2 alpha = (1-cos2alpha)/2$

${cos}^{2} α = \frac{1 + cos 2 α}{2}$ $cos^2 alpha = (1+cos2alpha)/2$

to have:

${sin}^{4} (2 x) = {({sin}^{2} (2 x))}^{2} = \frac{{(1 - cos (4 x))}^{2}}{4}$ $sin^4(2x) = (sin^2(2x))^2 = (1-cos(4x))^2/4$

${sin}^{4} (2 x) = \frac{1 - 2 cos (4 x) + {cos}^{2} (4 x)}{4}$ $sin^4(2x) = (1-2cos(4x) +cos^2(4x))/4$

${sin}^{4} (2 x) = \frac{1}{4} - \frac{cos (4 x)}{2} + \frac{1}{8} + \frac{cos (8 x)}{8}$ $sin^4(2x) = 1/4-(cos(4x))/2 +1/8 +(cos(8x))/8$

${sin}^{4} (2 x) = \frac{3}{8} - \frac{cos (4 x)}{2} + \frac{cos (8 x)}{8}$ $sin^4(2x) = 3/8-(cos(4x))/2 +(cos(8x))/8$

Using the linearity of the integral:

$\int {sin}^{4} (2 x) d x = \frac{3}{8} \int d x - \frac{1}{2} \int cos (4 x) d x + \frac{1}{8} \int cos (8 x) d x$ $int sin^4(2x) dx= 3/8int dx-1/2 int cos(4x)dx +1/8 intcos(8x)dx$

$\int {sin}^{4} (2 x) d x = \frac{3 x}{8} - \frac{1}{8} \int cos (4 x) d (4 x) + \frac{1}{64} \int cos (8 x) d (8 x)$ $int sin^4(2x) dx= (3x)/8-1/8 int cos(4x)d(4x) +1/64 intcos(8x)d(8x)$

$\int {sin}^{4} (2 x) d x = \frac{3 x}{8} - \frac{sin (4 x)}{8} + \frac{sin (8 x)}{64} + C$ $int sin^4(2x) dx= (3x)/8-(sin(4x))/8 + (sin(8x))/64+C$

Answer 2 · 2018-04-24T07:11:40

The answer is $= \frac{1}{64} sin (8 x) - \frac{1}{8} sin (4 x) + \frac{3}{8} x + C$ $=1/64sin(8x)-1/8sin(4x)+3/8x+C$

Explanation:

Apply Euler's identity

$sin (2 x) = \frac{e^{2 i x} - e^{- 2 i x}}{2 i}$ $sin(2x)=(e^(2ix)-e^(-2ix))/(2i)$

$e^{i x} = cos x + i sin x$ $e^(ix)=cosx+isinx$

Therefore,

${sin}^{4} (2 x) = {(\frac{e^{2 i x} - e^{- 2 i x}}{2 i})}^{4}$ $sin^4(2x)=((e^(2ix)-e^(-2ix))/(2i))^4$

$= \frac{1}{16} ((e^{i 8 x} + e^{- i 8 x}) - 4 (e^{i 4 x} + e^{- i 4 x}) + 6)$ $=1/16((e^(i8x)+e^(-i8x))-4(e^(i4x)+e^(-i4x))+6)$

$= \frac{1}{8} (\frac{e^{i 8 x} + e^{- i 8 x}}{2} - 4 \frac{e^{i 4 x} + e^{- i 4 x}}{2} + \frac{6}{2})$ $=1/8((e^(i8x)+e^(-i8x))/2-4(e^(i4x)+e^(-i4x))/2+6/2)$

$= \frac{1}{8} cos (8 x) - \frac{1}{2} cos (4 x) + \frac{3}{8}$ $=1/8cos(8x)-1/2cos(4x)+3/8$

So,

$\int {sin}^{4} (2 x) d x = \int (\frac{1}{8} cos (8 x) - \frac{1}{2} cos (4 x) + \frac{3}{8}) d x$ $intsin^4(2x)dx=int(1/8cos(8x)-1/2cos(4x)+3/8)dx$

$= \frac{1}{64} sin (8 x) - \frac{1}{8} sin (4 x) + \frac{3}{8} x + C$ $=1/64sin(8x)-1/8sin(4x)+3/8x+C$

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