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$$\int \frac{dx}{4+3\sin (2x)}$$

$u=2x$

$du=2dx$

$$\frac{1}{2}\int \frac{du}{4+3\sin(u)}$$

$v=\tan(\frac{u}{2})$

$du=\frac{2dv}{1+v^2}$

\begin{align*} \frac{1}{2}\int \frac{du}{4+3\sin(u)}={}&\frac{1}{2}\int \frac{\frac{2dv}{1+v^2}}{4+\frac{6v}{1+v^2}}=\frac{1}{2}\int \frac{\frac{2dv}{1+v^2}}{\frac{4v^2+6v+4}{1+v^2}}={} \\ {}={}&\frac{1}{2}\int \frac{2dv}{4v^2+6v+4}=\frac{1}{2}\int \frac{2dv}{4v^2+6v+4}={} \\ {}={}&\int \frac{2dv}{4v^2+6v+4}=\int \frac{dv}{4(v+\frac{3}{4})^2-1} \end{align*}

$m=v+\frac{3}{4}$

$dm=dv$

\begin{align*} \int \frac{dm}{4m^2-1}={}&4\int \frac{dm}{m^2-\frac{1}{4}}=4\int \frac{dm}{m^2-(\frac{1}{2})^2}={} \\ {}={}&\frac{1}{2}\ln\left(\frac{m-\frac{1}{2}}{m+\frac{1}{2}}\right)+C=\frac{1}{2}\ln\left(\frac{v+\frac{3}{4}-\frac{1}{2}}{v+\frac{3}{4}+\frac{1}{2}}\right)+C={} \\ {}={}&\frac{1}{2}\ln\left(\frac{v-\frac{5}{4}}{v+\frac{11}{4}}\right)+C\frac{1}{2}\ln\left(\frac{v-\frac{5}{4}}{v+\frac{11}{4}}\right)+C={} \\ {}={}&\frac{1}{2}\ln\left(\frac{\tan(x)+\frac{5}{4}}{\tan(x)+\frac{11}{4}}\right)+C \end{align*}

I think I got it wrong, but I cannot find the mistake.

Michael Hardy
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gbox
  • 12,867

1 Answers1

2

$\frac{1}{2}\int \frac{\frac{2dv}{1+v^2}}{4+\frac{6v}{1+v^2}}=\frac{1}{2}\int \frac{\frac{2dv}{1+v^2}}{\frac{4v^2+6v+4}{1+v^2}}=\frac{1}{2}\int \frac{2dv}{4v^2+6v+4}=\frac{1}{2}\int \frac{2dv}{4v^2+6v+4}=\int \frac{2dv}{4v^2+6v+4}=\int \frac{dv}{4(v+\frac{3}{4})^2+\frac{7}{4}}$

Edit - You completed the square wrong. It should be $4(v+\frac{3}{4})^2+\frac{7}{4}$ not $4(v+\frac{3}{4})^2-1$

Jackh
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