What is the integral of (cos(6x))/(cos^2(3x)) ?

The integral of (cos(6x))/(cos^2(3x)) is 1/3 cos(6x)tan(3x)+1/3 (6x-sin(6x))+C

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integral of (cos(6x))/(cos^2(3x))

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

Simplify

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

Add a constant to the solution

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

What is the integral of (cos(6x))/(cos^2(3x)) ?
The integral of (cos(6x))/(cos^2(3x)) is 1/3 cos(6x)tan(3x)+1/3 (6x-sin(6x))+C