What is the integral of sec^2(3x)csc(3x) ?

The integral of sec^2(3x)csc(3x) is 1/3 (sec(3x)+ln|tan((3x)/2)|)+C

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integral of sec^2(3x)csc(3x)

\int sec^{2} (3 x) csc (3 x) d x

\frac{1}{3} (sec (3 x) + ln tan (\frac{3 x}{2})) + C

Solution steps

\int sec^{2} (3 x) csc (3 x) d x

Rewrite using trig identities

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

Apply u-substitution

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

Rewrite using trig identities

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

Apply Integration By Parts

= \frac{1}{3} (sec (u) - \int - cot (u) csc (u) tan (u) d u)

\int - cot (u) csc (u) tan (u) d u = - ln tan (\frac{u}{2})

= \frac{1}{3} (sec (u) - (- ln tan (\frac{u}{2})))

Substitute back

u = 3 x

= \frac{1}{3} (sec (3 x) - (- ln tan (\frac{3 x}{2})))

Simplify

= \frac{1}{3} (sec (3 x) + ln tan (\frac{3 x}{2}))

Add a constant to the solution

= \frac{1}{3} (sec (3 x) + ln tan (\frac{3 x}{2})) + C

What is the integral of sec^2(3x)csc(3x) ?
The integral of sec^2(3x)csc(3x) is 1/3 (sec(3x)+ln|tan((3x)/2)|)+C