What is the integral of (1+sin(3x))/(cos^2(3x)) ?

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

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integral of (1+sin(3x))/(cos^2(3x))

\int \frac{1 + sin ( 3 x )}{cos ^{2} ( 3 x )} d x

\frac{1}{3} (tan (3 x) + sec (3 x)) + C

Solution steps

\int \frac{1 + sin ( 3 x )}{cos ^{2} ( 3 x )} d x

Apply u-substitution

: \int \frac{1 + sin ( u )}{3 cos ^{2} ( u )} d u

= \int \frac{1 + sin ( u )}{3 cos ^{2} ( u )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{1 + sin ( u )}{cos ^{2} ( u )} d u

Expand

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

= \frac{1}{3} \cdot \int \frac{1}{cos ^{2} ( u )} + \frac{sin ( u )}{cos ^{2} ( u )} d u

Apply the Sum Rule:

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

= \frac{1}{3} (\int \frac{1}{cos ^{2} ( u )} d u + \int \frac{sin ( u )}{cos ^{2} ( u )} d u)

\int \frac{1}{cos ^{2} ( u )} d u = tan (u)

\int \frac{sin ( u )}{cos ^{2} ( u )} d u = sec (u)

= \frac{1}{3} (tan (u) + sec (u))

Substitute back

u = 3 x

= \frac{1}{3} (tan (3 x) + sec (3 x))

Add a constant to the solution

= \frac{1}{3} (tan (3 x) + sec (3 x)) + C

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