Grade 12Integral Calculus

evaluate the following integral
∫ tan 2x tan 3x tan 5x dx

taniska

11 Years agoGrade 12

1 Answer

Jitender Singh

11 Years ago

Ans:
Hello Student,
Please find answer to your question below

$I = \int tan(2x)tan(3x)tan(5x)dx$
$tan5x = tan(3x+2x) = \frac{ tan3x+tan2x}{1-tan(3x)tan(2x)}$
$tan3x+tan2x = tan5x-tan(2x)tan(3x)tan(5x)$
Put in the integral
$I = \int [tan5x - tan3x - tan2x]dx$
$I = \int tan5xdx - \int tan3xdx - \int tan2xdx$
$I = \frac{log(sec5x)}{5} - \int tan3xdx - \int tan2xdx$
$I = \frac{log(sec5x)}{5} - \frac{log(sec3x)}{3} - \int tan2xdx$
$I = \frac{log(sec5x)}{5} - \frac{log(sec3x)}{3} - \frac{log(sec2x)}{2}+c$