What is the integral of sec^3(5x)tan^2(5x) ?

The integral of sec^3(5x)tan^2(5x) is 1/40 (-sec(5x)tan(5x)-ln|tan(5x)+sec(5x)|+2sec^4(5x)sin(5x))+C

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integral of sec^3(5x)tan^2(5x)

\int sec^{3} (5 x) tan^{2} (5 x) d x

\frac{1}{40} (- sec (5 x) tan (5 x) - ln ∣ tan (5 x) + sec (5 x) ∣ + 2 sec^{4} (5 x) sin (5 x)) + C

Solution steps

\int sec^{3} (5 x) tan^{2} (5 x) d x

Apply u-substitution

: \int sec^{3} (u) tan^{2} (u) \frac{1}{5} d u

= \int sec^{3} (u) tan^{2} (u) \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int sec^{3} (u) tan^{2} (u) d u

Rewrite using trig identities

= \frac{1}{5} \cdot \int sec^{3} (u) (- 1 + sec^{2} (u)) d u

Expand

sec^{3} (u) (- 1 + sec^{2} (u)) : - sec^{3} (u) + sec^{5} (u)

= \frac{1}{5} \cdot \int - sec^{3} (u) + sec^{5} (u) d u

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}{5} (- \int sec^{3} (u) d u + \int sec^{5} (u) d u)

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

\int sec^{5} (u) d u = \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

= \frac{1}{5} (- (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣) + \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣))

Substitute back

u = 5 x

= \frac{1}{5} (- (\frac{1}{2} sec (5 x) tan (5 x) + \frac{1}{2} ln ∣ tan (5 x) + sec (5 x) ∣) + \frac{sec ^{4} ( 5 x ) sin ( 5 x )}{4} + \frac{3}{4} (\frac{1}{2} sec (5 x) tan (5 x) + \frac{1}{2} ln ∣ tan (5 x) + sec (5 x) ∣))

Simplify

\frac{1}{5} (- (\frac{1}{2} sec (5 x) tan (5 x) + \frac{1}{2} ln ∣ tan (5 x) + sec (5 x) ∣) + \frac{sec ^{4} ( 5 x ) sin ( 5 x )}{4} + \frac{3}{4} (\frac{1}{2} sec (5 x) tan (5 x) + \frac{1}{2} ln ∣ tan (5 x) + sec (5 x) ∣)) : \frac{1}{40} (- sec (5 x) tan (5 x) - ln ∣ tan (5 x) + sec (5 x) ∣ + 2 sec^{4} (5 x) sin (5 x))

= \frac{1}{40} (- sec (5 x) tan (5 x) - ln ∣ tan (5 x) + sec (5 x) ∣ + 2 sec^{4} (5 x) sin (5 x))

Add a constant to the solution

= \frac{1}{40} (- sec (5 x) tan (5 x) - ln ∣ tan (5 x) + sec (5 x) ∣ + 2 sec^{4} (5 x) sin (5 x)) + C

\frac{d y}{d x} = y

\frac{d}{d x} (e^{x^{2} - 5})

\frac{\partial}{\partial x} (\frac{m}{x})

\int \frac{1}{( x + 1 ) ^{2}} d x

f (x) = \frac{9}{8 - x}

What is the integral of sec^3(5x)tan^2(5x) ?
The integral of sec^3(5x)tan^2(5x) is 1/40 (-sec(5x)tan(5x)-ln|tan(5x)+sec(5x)|+2sec^4(5x)sin(5x))+C