What is the integral of sec^2(5x)tan^4(5x) ?

The integral of sec^2(5x)tan^4(5x) is 1/25 tan^5(5x)+C

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integral of sec^2(5x)tan^4(5x)

\int sec^{2} (5 x) tan^{4} (5 x) d x

\frac{1}{25} tan^{5} (5 x) + C

Solution steps

\int sec^{2} (5 x) tan^{4} (5 x) d x

Apply u-substitution

: \int sec^{2} (u) tan^{4} (u) \frac{1}{5} d u

= \int sec^{2} (u) tan^{4} (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^{2} (u) tan^{4} (u) d u

Rewrite using trig identities

= \frac{1}{5} \cdot \int (1 + tan^{2} (u)) tan^{4} (u) d u

Apply u-substitution

: \int v^{4} d v

= \frac{1}{5} \cdot \int v^{4} d v

Apply the Power Rule

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

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

Substitute back

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

Simplify

\frac{1}{5} \cdot \frac{tan ^{5} ( 5 x )}{5} : \frac{1}{25} tan^{5} (5 x)

= \frac{1}{25} tan^{5} (5 x)

Add a constant to the solution

= \frac{1}{25} tan^{5} (5 x) + C

x \to 1 lim (\frac{x ^{2} - 4}{x - 1})

f (t) = e^{t + 7}

3 y

\frac{d}{d x} (\frac{1}{3 x ^{2}} - \frac{5}{2 x} + 8 x - 6)

\frac{d}{d x} (2 cos (- x))

What is the integral of sec^2(5x)tan^4(5x) ?
The integral of sec^2(5x)tan^4(5x) is 1/25 tan^5(5x)+C