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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{x ^{3} + x ^{2}}{x + 2} d x

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\int \frac{- 1}{x ^{2} + 4} d x

\int \frac{x ^{4}}{( 1 + x ^{2} )} d x

\int sec (x) tan (x) cos (sec (x)) d x

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