What is the integral of sec^5(7x)tan(7x) ?

The integral of sec^5(7x)tan(7x) is 1/35 sec^5(7x)+C

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integral of sec^5(7x)tan(7x)

\int sec^{5} (7 x) tan (7 x) d x

\frac{1}{35} sec^{5} (7 x) + C

Solution steps

\int sec^{5} (7 x) tan (7 x) d x

Apply u-substitution

: \int sec^{5} (u) tan (u) \frac{1}{7} d u

= \int sec^{5} (u) tan (u) \frac{1}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int sec^{5} (u) tan (u) d u

Apply u-substitution

: \int v^{4} d v

= \frac{1}{7} \cdot \int v^{4} d v

Apply the Power Rule

: \frac{v ^{5}}{5}

= \frac{1}{7} \cdot \frac{v ^{5}}{5}

Substitute back

= \frac{1}{7} \cdot \frac{sec ^{5} ( 7 x )}{5}

Simplify

\frac{1}{7} \cdot \frac{sec ^{5} ( 7 x )}{5} : \frac{1}{35} sec^{5} (7 x)

= \frac{1}{35} sec^{5} (7 x)

Add a constant to the solution

= \frac{1}{35} sec^{5} (7 x) + C

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What is the integral of sec^5(7x)tan(7x) ?
The integral of sec^5(7x)tan(7x) is 1/35 sec^5(7x)+C