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Question Number 152841 by 0731619 last updated on 02/Sep/21

Answered by bobhans last updated on 02/Sep/21

$$\left(2\right)\frac{1}{4}\int \sin {}^2\left(2x\right)dx=$$$$\frac{1}{4}\int \frac{1-\cos \left(4x\right)}{2}dx=$$$$\frac{1}{4}(\frac{x}{2}-\frac{1}{8}\sin \left(4x\right))+c=$$$$\frac{x}{8}-\frac{\sin \left(4x\right)}{32}+c$$

Answered by bobhans last updated on 02/Sep/21

$$\left(1\right)\int \frac{2\sin 2\theta }{\cos \left(2\theta \right)\sqrt{1+3\cos {}^2\left(2\theta \right)}}d\theta =$$$$\int \frac{-d\left(\cos \left(2\theta \right)\right)}{\cos \left(2\theta \right)\sqrt{1+3\cos {}^2\left(2\theta \right)}}=$$$$\int \frac{-du}{u\sqrt{1+3u^2}}=\int \frac{-1}{u}du+\int \frac{1}{\sqrt{1+3u^2}}du$$$$=-\ln \mid u\mid +\int \frac{du}{\sqrt{1+{\left(u\sqrt{3}\right)}^2}}$$$$=-\ln \mid \cos 2\left(\theta \right)\mid +\frac{1}{\sqrt{3}}\ln \mid \sqrt{1+3\cos {}^2\left(2\theta \right)}\sqrt{3}\cos \left(2\theta \right)\mid +c$$$$=\ln \mid \sec \left(2\theta \right)\mid +\frac{1}{\sqrt{3}}\ln \mid \cos \left(2\theta \right)\sqrt{3+9\cos {}^2\left(2\theta \right)}\mid +c$$

Commented by peter frank last updated on 03/Sep/21

$$\mathrm{good}$$

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