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Evaluate the indefinite integral $$\int \sin 2x \sqrt{\cos2x+1}\ dx$$

Hello, I am a Calc I student currently working on substitution, and cannot find a solution to this particular problem. Thank you for your time!

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    $\begingroup$ Substitution, let $u=\cos(2x)+1$. $\endgroup$ – André Nicolas Dec 7 '15 at 8:06
  • $\begingroup$ Welcome to MSE, please us MathJax when asking your question. $\endgroup$ – Nizar Dec 7 '15 at 8:08
  • $\begingroup$ Great question! $\endgroup$ – Mone Skratt Henry Dec 9 '15 at 13:22
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$$\int\sin(2x)\sqrt{\cos(2x)+1}\space\text{d}x=$$


Substitute $u=2x$ and $\text{d}u=2\space\text{d}x$:


$$\frac{1}{2}\int\sin(u)\sqrt{\cos(u)+1}\space\text{d}u=$$


Substitute $s=\cos(u)+1$ and $\text{d}s=-\sin(u)\space\text{d}u$:


$$-\frac{1}{2}\int\sqrt{s}\space\text{d}s=$$ $$-\frac{1}{2}\int s^{\frac{1}{2}}\space\text{d}s=$$ $$-\frac{1}{2}\cdot\frac{2s^{\frac{3}{2}}}{3}+\text{C}=$$ $$-\frac{s^{\frac{3}{3}}}{2}+\text{C}=$$ $$-\frac{(\cos(u)+1)^{\frac{3}{3}}}{2}+\text{C}=$$ $$-\frac{(\cos(2x)+1)^{\frac{3}{3}}}{2}+\text{C}=-\frac{\left(1+\cos(2x)\right)^{\frac{3}{2}}}{3}+\text{C}$$

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$u = \cos(2x) + 1$

Then we have

$$\int \sqrt{u} \frac{-1}{2} du$$

because

$$\frac{-1}{2} du = \sin(2x) dx$$


$$\int \sqrt{u} \frac{-1}{2} du$$

$$ = {u}^{3/2} \frac{-2}{2(3)} + C$$

$$ = {u}^{3/2} \frac{-1}{3} + C$$

$$ = \frac{-{u}^{3/2}}{3} + C$$

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