What is the integral of (1+sin(2x))/(cos(2x)) ?

The integral of (1+sin(2x))/(cos(2x)) is 1/2 ln|tan(2x)+sec(2x)|-1/2 ln|cos(2x)|+C

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integral of (1+sin(2x))/(cos(2x))

\int \frac{1 + sin ( 2 x )}{cos ( 2 x )} d x

\frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣ - \frac{1}{2} ln ∣ cos (2 x) ∣ + C

Solution steps

\int \frac{1 + sin ( 2 x )}{cos ( 2 x )} d x

Expand

\frac{1 + sin ( 2 x )}{cos ( 2 x )} : \frac{1}{cos ( 2 x )} + \frac{sin ( 2 x )}{cos ( 2 x )}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{cos ( 2 x )} d x = \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣

\int \frac{sin ( 2 x )}{cos ( 2 x )} d x = - \frac{1}{2} ln ∣ cos (2 x) ∣

= \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣ - \frac{1}{2} ln ∣ cos (2 x) ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣ - \frac{1}{2} ln ∣ cos (2 x) ∣ + C

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What is the integral of (1+sin(2x))/(cos(2x)) ?
The integral of (1+sin(2x))/(cos(2x)) is 1/2 ln|tan(2x)+sec(2x)|-1/2 ln|cos(2x)|+C