What is the integral of csc^4(5x)cot^4(5x) ?

The integral of csc^4(5x)cot^4(5x) is -1/5 (1/7 cot^7(5x)+1/5 cot^5(5x))+C

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integral of csc^4(5x)cot^4(5x)

\int csc^{4} (5 x) cot^{4} (5 x) d x

- \frac{1}{5} (\frac{1}{7} cot^{7} (5 x) + \frac{1}{5} cot^{5} (5 x)) + C

Solution steps

\int csc^{4} (5 x) cot^{4} (5 x) d x

Apply u-substitution

: \int csc^{4} (u) cot^{4} (u) \frac{1}{5} d u

= \int csc^{4} (u) cot^{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 csc^{4} (u) cot^{4} (u) d u

Apply Trigonometric Substitution

: \int - v^{4} (v^{2} + 1) d v

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

Take the constant out:

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

= \frac{1}{5} (- \int v^{4} (v^{2} + 1) d v)

Expand

v^{4} (v^{2} + 1) : v^{6} + v^{4}

= \frac{1}{5} (- \int v^{6} + 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}{5} (- (\int v^{6} d v + \int v^{4} d v))

\int v^{6} d v = \frac{v ^{7}}{7}

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

= \frac{1}{5} (- (\frac{v ^{7}}{7} + \frac{v ^{5}}{5}))

Substitute back

= \frac{1}{5} (- (\frac{cot ^{7} ( 5 x )}{7} + \frac{cot ^{5} ( 5 x )}{5}))

Simplify

\frac{1}{5} (- (\frac{cot ^{7} ( 5 x )}{7} + \frac{cot ^{5} ( 5 x )}{5})) : - \frac{1}{5} (\frac{1}{7} cot^{7} (5 x) + \frac{1}{5} cot^{5} (5 x))

= - \frac{1}{5} (\frac{1}{7} cot^{7} (5 x) + \frac{1}{5} cot^{5} (5 x))

Add a constant to the solution

= - \frac{1}{5} (\frac{1}{7} cot^{7} (5 x) + \frac{1}{5} cot^{5} (5 x)) + C

\frac{d}{d x} (2 x e^{- x^{2}})

(y^{16}) d y = \frac{( 1 + x )}{x} d x, y (1) = 5

\frac{( s ^{2} + s )}{2 s ^{2} + 2 s + 1}

f (x) = \frac{1}{( 3 x + 8 ) ^{2}}

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

What is the integral of csc^4(5x)cot^4(5x) ?
The integral of csc^4(5x)cot^4(5x) is -1/5 (1/7 cot^7(5x)+1/5 cot^5(5x))+C