What is the integral of 5sec^2(5x)tan^5(5x) ?

The integral of 5sec^2(5x)tan^5(5x) is 1/6 tan^6(5x)+C

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integral of 5sec^2(5x)tan^5(5x)

\int 5 sec^{2} (5 x) tan^{5} (5 x) d x

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

Solution steps

\int 5 sec^{2} (5 x) tan^{5} (5 x) d x

Take the constant out:

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

= 5 \cdot \int sec^{2} (5 x) tan^{5} (5 x) d x

Apply u-substitution

: \int sec^{2} (u) tan^{5} (u) \frac{1}{5} d u

= 5 \cdot \int sec^{2} (u) tan^{5} (u) \frac{1}{5} d u

Take the constant out:

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

= 5 \cdot \frac{1}{5} \cdot \int sec^{2} (u) tan^{5} (u) d u

Rewrite using trig identities

= 5 \cdot \frac{1}{5} \cdot \int (1 + tan^{2} (u)) tan^{5} (u) d u

Apply u-substitution

: \int v^{5} d v

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

Apply the Power Rule

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{( x + 1 )}{2})

\frac{d}{d t} (a (1 - cos (t)))

\frac{d y}{d t} = e^{t} (y - 1)^{2}

ln (3 x) - \frac{1}{6} x

x \to 2 lim (\frac{x + 2}{x ^{2} - 4})

What is the integral of 5sec^2(5x)tan^5(5x) ?
The integral of 5sec^2(5x)tan^5(5x) is 1/6 tan^6(5x)+C