What is the integral of 1/(cot^2(8x))sec^4(8x) ?

The integral of 1/(cot^2(8x))sec^4(8x) is 1/8 ((tan^3(8x))/3+(tan^5(8x))/5)+C

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integral of 1/(cot^2(8x))sec^4(8x)

\int \frac{1}{cot ^{2} ( 8 x )} sec^{4} (8 x) d x

\frac{1}{8} (\frac{tan ^{3} ( 8 x )}{3} + \frac{tan ^{5} ( 8 x )}{5}) + C

Solution steps

\int \frac{1}{cot ^{2} ( 8 x )} sec^{4} (8 x) d x

Simplify

\frac{1}{cot ^{2} ( 8 x )} sec^{4} (8 x) : sec^{4} (8 x) tan^{2} (8 x)

= \int sec^{4} (8 x) tan^{2} (8 x) d x

Apply u-substitution

: \int sec^{4} (u) tan^{2} (u) \frac{1}{8} d u

= \int sec^{4} (u) tan^{2} (u) \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int sec^{4} (u) tan^{2} (u) d u

Rewrite using trig identities

= \frac{1}{8} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) tan^{2} (u) d u

Apply u-substitution

: \int v^{2} (1 + v^{2}) d v

= \frac{1}{8} \cdot \int v^{2} (1 + v^{2}) d v

Expand

v^{2} (1 + v^{2}) : v^{2} + v^{4}

= \frac{1}{8} \cdot \int v^{2} + v^{4} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (\int v^{2} d v + \int v^{4} d v)

\int v^{2} d v = \frac{v ^{3}}{3}

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

= \frac{1}{8} (\frac{v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= \frac{1}{8} (\frac{tan ^{3} ( 8 x )}{3} + \frac{tan ^{5} ( 8 x )}{5})

Add a constant to the solution

= \frac{1}{8} (\frac{tan ^{3} ( 8 x )}{3} + \frac{tan ^{5} ( 8 x )}{5}) + C

\frac{\partial}{\partial x} (2 x^{2} y cos (x))

\frac{d}{d x} (1 - x^{2} - y^{2} - z^{2})

\int (3 x + 3) x^{2} + 2 x + 3 d x

n = 1 \sum \infty 9 - \frac{1}{n ^{2}}

n = 1 \sum \infty 10 (- \frac{2}{3})^{n}

What is the integral of 1/(cot^2(8x))sec^4(8x) ?
The integral of 1/(cot^2(8x))sec^4(8x) is 1/8 ((tan^3(8x))/3+(tan^5(8x))/5)+C