Question

Please help... Integrate sqrt(1-cos3x)

Answer 1

\[\text{ Recall } \sqrt{\frac{1+\cos(x)}{2}}=\cos(\frac{x}{2}) \\ \sqrt{1+\cos(x)}=\sqrt{2} \cos(\frac{x}{2})\]

Answer 2

actually equal to plus or minus

Answer 3

doesn't affect that it's 3x?

Answer 4

\[\text{ why don't you replace x with 3x }\]

Answer 5

another way:
\[\int\limits_{}^{}\sqrt{1-\cos(3x)} \cdot \frac{\sqrt{1+\cos(3x)}}{\sqrt{1+\cos(3x)}} dx\\ \int\limits_{}^{}\frac{\sqrt{1-\cos^2(3x)}}{\sqrt{1+\cos(3x)}} dx \\ \int\limits_{}^{} \frac{\sqrt{\sin^2(3x)}}{\sqrt{1+\cos(3x)}} \\ \text{ assume } \sin(3x)>0 \\ \int\limits_{}^{}\frac{\sin(3x)}{\sqrt{1+\cos(3x)}} dx \\ \text{ let } u=1+\cos(3x)\]

Answer 6

I see... I think I've finish but when I replace x with 2pi/3, my result it's 0 and it's different from the result I must have. My integral it's from 0 to 2pi/3 and I alredy have 16/3...
\[\frac{ 16 a}{ 3 } (\sin \frac{ 3x }{ 2 })\]

Answer 7

\[\sqrt{1-\cos(3x)}=\sqrt{2} \sin(\frac{3x}{2}) \\ \int\limits_{0}^{\frac{2\pi}{3}}\sqrt{2} \sin(\frac{3x}{2}) dx=-\sqrt{2} \cdot \frac{2}{3} \cos(\frac{3x}{2})|_0^{\frac{2 \pi}{3}}=\]

Answer 8

Thank you @freckles , you helped me a lot! I was replacing incorrectly but now I see all clear :)!

Answer 9

Well the identity I gave you above (the first post) was the half angle identity for cosine
Was hoping it would be a hint to use the half angle identity for sine :p