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

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

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

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

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

Solution steps

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

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

= \int \frac{tan ( 3 x )}{cos ( 3 x )} d x

Apply u-substitution

= \int \frac{tan ( u )}{3 cos ( 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{tan ( u )}{cos ( u )} d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int tan (u) sec (u) d u

Apply u-substitution

= \frac{1}{3} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot v

Substitute back

= \frac{1}{3} \cdot 1 \cdot sec (3 x)

Simplify

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

Add a constant to the solution

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

a re a 2 x^{2}, 2 x + 2

\int e^{x^{2} - x} d x

\frac{\partial}{\partial x} (arccos (\frac{x}{x ^{2} + y ^{2}}))

7 y^{^{'}} = y^{2}

d er i v a t i v e f (x) = x e^{3 x^{2} + 1}

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