What is the integral of 7sec^3(2x) ?

The integral of 7sec^3(2x) is 7/2 (1/2 sec(2x)tan(2x)+1/2 ln|tan(2x)+sec(2x)|)+C

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integral of 7sec^3(2x)

\int 7 sec^{3} (2 x) d x

\frac{7}{2} (\frac{1}{2} sec (2 x) tan (2 x) + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣) + C

Solution steps

\int 7 sec^{3} (2 x) d x

Take the constant out:

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

= 7 \cdot \int sec^{3} (2 x) d x

Apply u-substitution

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

= 7 \cdot \int sec^{3} (u) \frac{1}{2} d u

Take the constant out:

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

= 7 \cdot \frac{1}{2} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 7 \cdot \frac{1}{2} (\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) ∣

= 7 \cdot \frac{1}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = 2 x

= 7 \cdot \frac{1}{2} (\frac{sec ^{2} ( 2 x ) sin ( 2 x )}{2} + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣)

Simplify

7 \cdot \frac{1}{2} (\frac{sec ^{2} ( 2 x ) sin ( 2 x )}{2} + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣) : \frac{7}{2} (\frac{1}{2} sec (2 x) tan (2 x) + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣)

= \frac{7}{2} (\frac{1}{2} sec (2 x) tan (2 x) + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣)

Add a constant to the solution

= \frac{7}{2} (\frac{1}{2} sec (2 x) tan (2 x) + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣) + C

\frac{d y}{d x} + y^{2} - y = 0

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y = ln (x^{3} sin^{2} (x))

What is the integral of 7sec^3(2x) ?
The integral of 7sec^3(2x) is 7/2 (1/2 sec(2x)tan(2x)+1/2 ln|tan(2x)+sec(2x)|)+C