I have tried to integrate the following indefinite integral but I'm not sure if I get the right answer. Please tell me if I'm wrong and if so, please indicate what went wrong.

$$ \int\sqrt{\cos2x}\sin^32x\,dx $$ $$ \int\sqrt{\cos2x}(\sin^22x)(\sin2x)\,dx $$ $$ \int\sqrt{\cos2x}(1-\cos^22x)(\sin2x)\,dx $$ $$ \frac {-1}2 \int\sqrt{u}(1-u^2)\,du $$ $$ \frac {-1}2 \int(u^{\frac 12}-u^{\frac 52})\,du $$ $$ \frac {-1}2 (\frac {2u^{\frac 32}}3-\frac {2u^{\frac 72}}7)\,+C $$ $$ \frac {u^{\frac 72}}7-\frac {u^{\frac 32}}3\,+C $$ $$ \frac {\sqrt{\cos^72x}}7-\frac {\sqrt{\cos^32x}}3\,+C $$

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    $\begingroup$ The answer is correct. Next time, it is a better idea to first check whether the result of the integral is correct using e.g. Wolfram|Alpha, Maple, or Mathematica. $\endgroup$
    – wythagoras
    Jul 29, 2015 at 18:35
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    $\begingroup$ The process is certainly right. The details also look right. $\endgroup$ Jul 29, 2015 at 18:35
  • $\begingroup$ I tried with Wolfram|Alpha before writing here, that's what made me wonder whether the result was correct. (wolframalpha.com/input/…) $\endgroup$
    – Mart
    Jul 29, 2015 at 18:44
  • $\begingroup$ Why not try some trig identities to show your answer is the same as WA's? $\endgroup$ Jul 29, 2015 at 19:48
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    $\begingroup$ You can check the correctness by differentiating the result. Differentiating your result gives the original function, so it's correct. $\endgroup$ Jul 30, 2015 at 0:09

1 Answer 1


Using the fact that $\cos^3 2x = \frac{\cos 6x+3 \cos 2x}{4}$ you get the answer given by Wolfram Alpha.


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