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

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

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

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

\frac{1}{6} sec (6 x) + C

Solution steps

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

Use the basic trigonometric identity:

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

= \int \frac{tan ( 6 x )}{cos ( 6 x )} d x

Apply u-substitution

: \int \frac{tan ( u )}{6 cos ( u )} d u

= \int \frac{tan ( u )}{6 cos ( u )} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{tan ( u )}{cos ( u )} d u

Rewrite using trig identities

= \frac{1}{6} \cdot \int tan (u) sec (u) d u

Apply u-substitution

: \int 1 d v

= \frac{1}{6} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{6} \cdot 1 \cdot v

Substitute back

= \frac{1}{6} \cdot 1 \cdot sec (6 x)

Simplify

= \frac{1}{6} sec (6 x)

Add a constant to the solution

= \frac{1}{6} sec (6 x) + C

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