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

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

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integral of sec^3(5x)

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integral Reduction

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

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

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

Substitute back

u = 5 x

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

Add a constant to the solution

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

y = arcsin (x)

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

2 z

\frac{\partial}{\partial y} (x^{2} \cdot e^{- 2 y})

\frac{d y}{d x}, y = x^{π}

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