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

The integral of (sec(3u)+csc(3u))^2 is 1/3 (tan(3u)+ln(tan^2(3u))-cot(3u))+C

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integral of (sec(3u)+csc(3u))^2

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

\frac{1}{3} (tan (3 u) + ln (tan^{2} (3 u)) - cot (3 u)) + C

Solution steps

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

Expand

(sec (3 u) + csc (3 u))^{2} : sec^{2} (3 u) + 2 sec (3 u) csc (3 u) + csc^{2} (3 u)

= \int sec^{2} (3 u) + 2 sec (3 u) csc (3 u) + csc^{2} (3 u) d u

Apply u-substitution

: \int (sec^{2} (v) + 2 sec (v) csc (v) + csc^{2} (v)) \frac{1}{3} d v

= \int (sec^{2} (v) + 2 sec (v) csc (v) + csc^{2} (v)) \frac{1}{3} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int sec^{2} (v) + 2 sec (v) csc (v) + csc^{2} (v) d v

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 sec^{2} (v) d v + \int 2 sec (v) csc (v) d v + \int csc^{2} (v) d v)

\int sec^{2} (v) d v = tan (v)

\int 2 sec (v) csc (v) d v = ln (tan^{2} (v))

\int csc^{2} (v) d v = - cot (v)

= \frac{1}{3} (tan (v) + ln (tan^{2} (v)) - cot (v))

Substitute back

v = 3 u

= \frac{1}{3} (tan (3 u) + ln (tan^{2} (3 u)) - cot (3 u))

Add a constant to the solution

= \frac{1}{3} (tan (3 u) + ln (tan^{2} (3 u)) - cot (3 u)) + C

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