What is the integral of sec^6(5x)tan(5x) ?

The integral of sec^6(5x)tan(5x) is 1/30 sec^6(5x)+C

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integral of sec^6(5x)tan(5x)

\int sec^{6} (5 x) tan (5 x) d x

\frac{1}{30} sec^{6} (5 x) + C

Solution steps

\int sec^{6} (5 x) tan (5 x) d x

Apply u-substitution

: \int sec^{6} (u) tan (u) \frac{1}{5} d u

= \int sec^{6} (u) tan (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^{6} (u) tan (u) d u

Apply u-substitution

: \int v^{5} d v

= \frac{1}{5} \cdot \int v^{5} d v

Apply the Power Rule

: \frac{v ^{6}}{6}

= \frac{1}{5} \cdot \frac{v ^{6}}{6}

Substitute back

= \frac{1}{5} \cdot \frac{sec ^{6} ( 5 x )}{6}

Simplify

\frac{1}{5} \cdot \frac{sec ^{6} ( 5 x )}{6} : \frac{1}{30} sec^{6} (5 x)

= \frac{1}{30} sec^{6} (5 x)

Add a constant to the solution

= \frac{1}{30} sec^{6} (5 x) + C

(1 - x^{2})^{^{'}}

y^{^{'}} + x y + 4 = 0

\int \frac{1}{x ^{2} - 36} d x

\int \frac{1}{8 + 6 x - 9 x ^{2}} d x

\frac{d}{d x} (f (h, x) (g (h, x) (h (x))))

What is the integral of sec^6(5x)tan(5x) ?
The integral of sec^6(5x)tan(5x) is 1/30 sec^6(5x)+C