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I'm stuck at this problem:

$$ \int{\sqrt{(\sin^2 x)^2 + (2\sin x \cos x)^2}dx} = \int{\sqrt{\sin^2 x \sin^2 x + 4\sin^2 x \cos^2 x} dx}$$

I tried a few trig identities: $\sin^2 x = \frac{1-\cos 2x}{2} $ and $ \cos^2 x = \frac{1+\cos 2x}{2} $ and $\sin^2 x + \cos^2 x = 1$. Keep hitting dead end. Any tips?

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  • $\begingroup$ The answer doesn't look pretty according to wolframalpha. $\endgroup$ – Ian Miller May 29 '16 at 16:28
  • $\begingroup$ Have you tried $\sin 2 x = 2 \sin x \cos x$ and $\cos 2 x = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x$? (Hoping it's not a proper elliptic integral...) $\endgroup$ – Andrew Miloradovsky May 29 '16 at 16:35
  • $\begingroup$ @Ian Miller I tried symbolab and cymath, both couldn't work it out. I think wolfram's legit. When I input the upper and lower bounds, it gives the answer same from the answer sheet. $ 4 + \frac{2}{\sqrt{3}} \ln (2+\sqrt{3}). $ wolframalpha.com/input/… $\endgroup$ – user321070 May 29 '16 at 16:48
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$$\begin{align} &\int{\sqrt{\sin^2x \sin^2x + 4\sin^2 x \cos^2 x} dx} \\ & =\int{\sqrt{\sin^2x( \sin^2x + 4\cos^2 x)} dx} \\ & =\begin{cases}\int{\sin x\sqrt{( \sin^2x + 4\cos^2 x)} dx} & & & &, \sin x \geq 0 \\ -\int{\sin x\sqrt{( \sin^2x + 4\cos^2 x)} dx} & & & &, \sin x < 0\end{cases} \\ & =\begin{cases}\int{\sin x\sqrt{(1 + 3\cos^2 x)} dx} & & & & & &, \sin x \geq 0 \\ -\int{\sin x\sqrt{(1 + 3\cos^2 x)} dx} & & & & & &, \sin x < 0\end{cases} \\ & =\begin{cases}-\frac{1}{\sqrt3}\int{\sin x\sqrt{(1 + 3\cos^2 x)} d(\sqrt3\cos x)} & , \sin x \geq 0 \\ \frac{1}{\sqrt3}\int{\sin x\sqrt{(1 + 3\cos^2 x)} d(\sqrt3\cos x)} & , \sin x < 0\end{cases}\end{align}$$

Now say $z=\sqrt3\cos x$. Then the integration becomes

$$=\begin{cases}-\frac{1}{\sqrt3}\int{\sin x\sqrt{1 + z^2} dz} & , \sin x \geq 0 \\ \frac{1}{\sqrt3}\int{\sin x\sqrt{1 + z^2} dz} & , \sin x < 0\end{cases}$$

I hope you can finish it now, using the necessary formula. The formula can be deduced using integration by parts and proper substitutions and is denoted in the linked page as no. 8.

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  • $\begingroup$ When you took the $\sin x$ out of the square root, wouldn't that be valid only where $\sin x\ge 0$? So perhaps a plus-or-minus or the equivalent should be added? $\endgroup$ – Rory Daulton May 29 '16 at 22:12
  • $\begingroup$ @RoryDaulton Provided appropriate changes. Thanks for the feedback. Please check if it is okay now. $\endgroup$ – SchrodingersCat May 30 '16 at 4:40
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Try a substitution with the last expression that you have

$$\int\sqrt{\sin^4(x)+4\sin^2(x)\cos^2(x)}dx\quad u=\cos(x)$$.

This leads us to the following integral where we do another substitution:

$$\int\sqrt{3u^2+1}du \quad\!\! \text{substitute}\quad\!\! u=\sqrt{\dfrac{a}{b}}\tan (v) \quad\!\!\text{where}\quad\!\! a=1 \quad\!\!\text{and} \quad\!\!b=3$$

Then, the integral becomes the following:

$$\int\dfrac{\sqrt{\tan^2(v)+1}+\sec^2(v)}{\sqrt{3}}$$

and the rest is more or less is using trig identities.

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  • $\begingroup$ I think you meant $\times \sec^2(v)$ rather than $+\sec^2(v)$. Also you're missing your $dv$ $\endgroup$ – Ian Miller May 29 '16 at 17:46

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