# Evaluating the indefinite integral $\int\sqrt{\cos2x}\sin^32x\,dx$

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.

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

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

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