How do I evaluate the indefinite integral $\int {sin}^{2} (2 t) d t$ $intsin^2(2t)dt$ ?

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How do I evaluate the indefinite integral $\int {sin}^{2} (2 t) d t$ $intsin^2(2t)dt$ ?

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Answer 1 · 2014-07-28T19:20:32

$= \frac{1}{2} t - \frac{1}{8} sin 4 t + c$ $=1/2t-1/8sin4t+c$ , where $c$ $c$ is a constant

Explanation,

$= \int {sin}^{2} (2 t) d t$ $=intsin^2(2t)dt$

Using trigonometric identity, $cos 2 t = 1 - 2 {sin}^{2} t$ $cos2t=1-2sin^2t$

${sin}^{2} t = \frac{1 - cos 2 t}{2}$ $sin^2t=(1-cos2t)/2$ , inserting this value of ${sin}^{2} t$ $sin^2t$ in integral, we get

$= \int \frac{1 - cos 4 t}{2} d t$ $=int(1-cos4t)/2dt$

$= \frac{1}{2} \int 1 d t - \frac{1}{2} \int cos 4 t d t$ $=1/2int1dt-1/2intcos4tdt$

$= \frac{1}{2} t - \frac{1}{2} \frac{sin 4 t}{4} + c$ $=1/2t-1/2(sin4t)/4+c$ , where $c$ $c$ is a constant

$= \frac{1}{2} t - \frac{1}{8} sin 4 t + c$ $=1/2t-1/8sin4t+c$ , where $c$ $c$ is a constant

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How do I evaluate the indefinite integral $\int {sin}^{2} (2 t) d t$ $intsin^2(2t)dt$ ?

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How do I evaluate the indefinite integral $\int {sin}^{2} (2 t) d t$ $intsin^2(2t)dt$ ?