What is the integral of 9sec^3(5x) ?

The integral of 9sec^3(5x) is 9/5 (1/2 sec^2(5x)sin(5x)+1/2 ln|tan(5x)+sec(5x)|)+C

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integral of 9sec^3(5x)

\int 9 sec^{3} (5 x) d x

\frac{9}{5} (\frac{1}{2} sec^{2} (5 x) sin (5 x) + \frac{1}{2} ln ∣ tan (5 x) + sec (5 x) ∣) + C

Solution steps

\int 9 sec^{3} (5 x) d x

Take the constant out:

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

= 9 \cdot \int sec^{3} (5 x) d x

Apply u-substitution

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

= 9 \cdot \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

= 9 \cdot \frac{1}{5} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 9 \cdot \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) ∣

= 9 \cdot \frac{1}{5} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = 5 x

= 9 \cdot \frac{1}{5} (\frac{sec ^{2} ( 5 x ) sin ( 5 x )}{2} + \frac{1}{2} ln ∣ tan (5 x) + sec (5 x) ∣)

Simplify

9 \cdot \frac{1}{5} (\frac{sec ^{2} ( 5 x ) sin ( 5 x )}{2} + \frac{1}{2} ln ∣ tan (5 x) + sec (5 x) ∣) : \frac{9}{5} (\frac{1}{2} sec^{2} (5 x) sin (5 x) + \frac{1}{2} ln ∣ tan (5 x) + sec (5 x) ∣)

= \frac{9}{5} (\frac{1}{2} sec^{2} (5 x) sin (5 x) + \frac{1}{2} ln ∣ tan (5 x) + sec (5 x) ∣)

Add a constant to the solution

= \frac{9}{5} (\frac{1}{2} sec^{2} (5 x) sin (5 x) + \frac{1}{2} ln ∣ tan (5 x) + sec (5 x) ∣) + C

8 x^{7} (x - 1)^{7} + 7 (x - 1)^{6} x^{8}

\frac{\partial}{\partial x} (7 y^{8} x^{6} - 6 y^{7} x^{5})

\int \frac{z ^{2}}{z + 1} d z

y^{^{''}} + y^{^{'}} + 3 y = 0, y (0) = 0, y^{^{'}} (0) = 0

\frac{3}{t}

What is the integral of 9sec^3(5x) ?
The integral of 9sec^3(5x) is 9/5 (1/2 sec^2(5x)sin(5x)+1/2 ln|tan(5x)+sec(5x)|)+C