What is the integral of sec(6x+3)tan(6x+3) ?

The integral of sec(6x+3)tan(6x+3) is 1/6 sin(6x+3)tan(6x+3)+1/6 cos(6x+3)+C

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integral of sec(6x+3)tan(6x+3)

\int sec (6 x + 3) tan (6 x + 3) d x

\frac{1}{6} sin (6 x + 3) tan (6 x + 3) + \frac{1}{6} cos (6 x + 3) + C

Solution steps

\int sec (6 x + 3) tan (6 x + 3) d x

Rewrite using trig identities

= \int \frac{sin ( 6 x + 3 )}{cos ^{2} ( 6 x + 3 )} d x

Apply Integration By Parts

: sin (6 x + 3) (- \frac{1}{2} + \frac{1}{6} tan (6 x + 3)) - \int - 3 cos (6 x + 3) + sin (3 + 6 x) d x

= sin (6 x + 3) (- \frac{1}{2} + \frac{1}{6} tan (6 x + 3)) - \int - 3 cos (6 x + 3) + sin (3 + 6 x) d x

\int - 3 cos (6 x + 3) + sin (3 + 6 x) d x = - \frac{1}{2} sin (6 x + 3) - \frac{1}{6} cos (3 + 6 x)

= sin (6 x + 3) (- \frac{1}{2} + \frac{1}{6} tan (6 x + 3)) - (- \frac{1}{2} sin (6 x + 3) - \frac{1}{6} cos (3 + 6 x))

Simplify

sin (6 x + 3) (- \frac{1}{2} + \frac{1}{6} tan (6 x + 3)) - (- \frac{1}{2} sin (6 x + 3) - \frac{1}{6} cos (3 + 6 x)) : \frac{1}{6} sin (6 x + 3) tan (6 x + 3) + \frac{1}{6} cos (6 x + 3)

= \frac{1}{6} sin (6 x + 3) tan (6 x + 3) + \frac{1}{6} cos (6 x + 3)

Add a constant to the solution

= \frac{1}{6} sin (6 x + 3) tan (6 x + 3) + \frac{1}{6} cos (6 x + 3) + C

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What is the integral of sec(6x+3)tan(6x+3) ?
The integral of sec(6x+3)tan(6x+3) is 1/6 sin(6x+3)tan(6x+3)+1/6 cos(6x+3)+C