What is the integral of sec^2(5x)tan^6(5x) ?

The integral of sec^2(5x)tan^6(5x) is 1/35 tan^7(5x)+C

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integral of sec^2(5x)tan^6(5x)

\int sec^{2} (5 x) tan^{6} (5 x) d x

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

Solution steps

\int sec^{2} (5 x) tan^{6} (5 x) d x

Apply u-substitution

: \int sec^{2} (u) tan^{6} (u) \frac{1}{5} d u

= \int sec^{2} (u) tan^{6} (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^{6} (u) d u

Rewrite using trig identities

= \frac{1}{5} \cdot \int (1 + tan^{2} (u)) tan^{6} (u) d u

Apply u-substitution

: \int v^{6} d v

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

Apply the Power Rule

: \frac{v ^{7}}{7}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

f (x) = x^{2} - 49 x - 456, at x = 2

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

\int (x^{2} + 8) e^{- x} d x

x \to \infty lim (\frac{6}{x})

y = 3 x + 1, (5, 16)

What is the integral of sec^2(5x)tan^6(5x) ?
The integral of sec^2(5x)tan^6(5x) is 1/35 tan^7(5x)+C