What is the integral of sec(3x+4)tan(3x+4) ?

The integral of sec(3x+4)tan(3x+4) is 1/3 sin(3x+4)tan(3x+4)+1/3 cos(3x+4)+C

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integral of sec(3x+4)tan(3x+4)

\int sec (3 x + 4) tan (3 x + 4) d x

\frac{1}{3} sin (3 x + 4) tan (3 x + 4) + \frac{1}{3} cos (3 x + 4) + C

Solution steps

\int sec (3 x + 4) tan (3 x + 4) d x

Rewrite using trig identities

= \int \frac{sin ( 3 x + 4 )}{cos ^{2} ( 3 x + 4 )} d x

Apply Integration By Parts

: sin (3 x + 4) (- \frac{4}{3} + \frac{1}{3} tan (3 x + 4)) - \int - 4 cos (3 x + 4) + sin (4 + 3 x) d x

= sin (3 x + 4) (- \frac{4}{3} + \frac{1}{3} tan (3 x + 4)) - \int - 4 cos (3 x + 4) + sin (4 + 3 x) d x

\int - 4 cos (3 x + 4) + sin (4 + 3 x) d x = - \frac{4}{3} sin (3 x + 4) - \frac{1}{3} cos (4 + 3 x)

= sin (3 x + 4) (- \frac{4}{3} + \frac{1}{3} tan (3 x + 4)) - (- \frac{4}{3} sin (3 x + 4) - \frac{1}{3} cos (4 + 3 x))

Simplify

sin (3 x + 4) (- \frac{4}{3} + \frac{1}{3} tan (3 x + 4)) - (- \frac{4}{3} sin (3 x + 4) - \frac{1}{3} cos (4 + 3 x)) : \frac{1}{3} sin (3 x + 4) tan (3 x + 4) + \frac{1}{3} cos (3 x + 4)

= \frac{1}{3} sin (3 x + 4) tan (3 x + 4) + \frac{1}{3} cos (3 x + 4)

Add a constant to the solution

= \frac{1}{3} sin (3 x + 4) tan (3 x + 4) + \frac{1}{3} cos (3 x + 4) + C

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What is the integral of sec(3x+4)tan(3x+4) ?
The integral of sec(3x+4)tan(3x+4) is 1/3 sin(3x+4)tan(3x+4)+1/3 cos(3x+4)+C