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Mathematics 17 Online
OpenStudy (anonymous):

Integrate please? (2r-1) cos sqrt[3(2r-1)^2 +6] / sqrt[3(2r-1)^2 +6]

OpenStudy (anonymous):

\[(2r-1) \cos \sqrt{3(2r-1)^{2}+ 6} / \sqrt{3(2r-1)^{2}+ 6}\]

OpenStudy (anonymous):

I think substituting the angel would be a nice start.

OpenStudy (anonymous):

3(2r-1)^2 +6 = t^2

OpenStudy (anonymous):

meaning?

OpenStudy (anonymous):

Integration by Substitution?

OpenStudy (anonymous):

yes, what ive got my final answer is 1/12 sin sqrt[3(2r-1)^2 +6] but its wrong :(

OpenStudy (anonymous):

\[(2r-1)\frac{\cos t}{t} \cdot\frac{2t}{12(2r-1)} dt\] \[\frac16\cdot \sin t\]

OpenStudy (anonymous):

or, 1/6 sin sqrt[3(2r-1)^2 +6]

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