What is the integral of 2sec^4(7x) ?

The integral of 2sec^4(7x) is 2/7 (tan(7x)+1/3 tan^3(7x))+C

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integral of 2sec^4(7x)

\int 2 sec^{4} (7 x) d x

\frac{2}{7} (tan (7 x) + \frac{1}{3} tan^{3} (7 x)) + C

Solution steps

\int 2 sec^{4} (7 x) d x

Take the constant out:

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

= 2 \cdot \int sec^{4} (7 x) d x

Apply u-substitution

: \int sec^{4} (u) \frac{1}{7} d u

= 2 \cdot \int sec^{4} (u) \frac{1}{7} d u

Take the constant out:

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

= 2 \cdot \frac{1}{7} \cdot \int sec^{4} (u) d u

Rewrite using trig identities

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

Apply u-substitution

: \int 1 + v^{2} d v

= 2 \cdot \frac{1}{7} \cdot \int 1 + v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot \frac{1}{7} (\int 1 d v + \int v^{2} d v)

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= 2 \cdot \frac{1}{7} (v + \frac{v ^{3}}{3})

Substitute back

= 2 \cdot \frac{1}{7} (tan (7 x) + \frac{tan ^{3} ( 7 x )}{3})

Simplify

2 \cdot \frac{1}{7} (tan (7 x) + \frac{tan ^{3} ( 7 x )}{3}) : \frac{2}{7} (tan (7 x) + \frac{1}{3} tan^{3} (7 x))

= \frac{2}{7} (tan (7 x) + \frac{1}{3} tan^{3} (7 x))

Add a constant to the solution

= \frac{2}{7} (tan (7 x) + \frac{1}{3} tan^{3} (7 x)) + C

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What is the integral of 2sec^4(7x) ?
The integral of 2sec^4(7x) is 2/7 (tan(7x)+1/3 tan^3(7x))+C