Solve the trigonometric integral int(arctan(3/5x))dx

 Exercise

$\int arc\:tan\:\left(\frac{3}{5}x\right)dx$

 

Apply integration by parts: $u=arctan(\frac{3}{5}x)$, $v'=1$

$x\arctan\left(\frac{3}{5}x\right)-\int\frac{\frac{3}{5}x}{1+\left(\frac{3}{5}x\right)^2}dx$

 

Simplify the expression

$x\arctan\left(\frac{3}{5}x\right)-\frac{3}{5}\int\frac{x}{1+\frac{9}{25}x^2}dx$

 

The integral $-\frac{3}{5}\int\frac{x}{1+\frac{9}{25}x^2}dx$ results in: $-\frac{5}{6}\ln\left(1+\frac{9}{25}x^2\right)$

$-\frac{5}{6}\ln\left(1+\frac{9}{25}x^2\right)$

 

Gather the results of all integrals

$x\arctan\left(\frac{3}{5}x\right)-\frac{5}{6}\ln\left|1+\frac{9}{25}x^2\right|$

 

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$x\arctan\left(\frac{3}{5}x\right)-\frac{5}{6}\ln\left|1+\frac{9}{25}x^2\right|+C_0$

$x\arctan\left(\frac{3}{5}x\right)-\frac{5}{6}\ln\left|1+\frac{9}{25}x^2\right|+C_0$

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