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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply Integration By Parts

: - sec (u) - \int - 2 sec (u) tan (u) d u

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

\int - 2 sec (u) tan (u) d u = - 2 sec (u)

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

Substitute back

u = 3 x

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

Simplify

\frac{1}{3} (- sec (3 x) - (- 2 sec (3 x))) : \frac{1}{3} sec (3 x)

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

Add a constant to the solution

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

\frac{d ^{2}}{d x ^{2}} (e^{- 4 x^{2}})

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

sec (4 x)

f (x) = cot (x)

\int 2 x + 8 e^{x} d x

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