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

The integral of 5csc^5(5x) is -1/4 csc^4(5x)cos(5x)+3/4 (-1/2 cot(5x)csc(5x)+1/2 ln|tan((5x)/2)|)+C

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integral of 5csc^5(5x)

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

- \frac{1}{4} csc^{4} (5 x) cos (5 x) + \frac{3}{4} (- \frac{1}{2} cot (5 x) csc (5 x) + \frac{1}{2} ln tan (\frac{5 x}{2})) + C

Solution steps

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

Apply u-substitution

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

= 5 \cdot \int csc^{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 csc^{5} (u) d u

Apply Integral Reduction

= 5 \cdot \frac{1}{5} (- \frac{cos ( u ) csc ^{4} ( u )}{4} + \frac{3}{4} \cdot \int csc^{3} (u) d u)

\int csc^{3} (u) d u = - \frac{1}{2} cot (u) csc (u) + \frac{1}{2} ln tan (\frac{u}{2})

= 5 \cdot \frac{1}{5} (- \frac{cos ( u ) csc ^{4} ( u )}{4} + \frac{3}{4} (- \frac{1}{2} cot (u) csc (u) + \frac{1}{2} ln tan (\frac{u}{2})))

Substitute back

u = 5 x

= 5 \cdot \frac{1}{5} (- \frac{cos ( 5 x ) csc ^{4} ( 5 x )}{4} + \frac{3}{4} (- \frac{1}{2} cot (5 x) csc (5 x) + \frac{1}{2} ln tan (\frac{5 x}{2})))

Simplify

5 \cdot \frac{1}{5} (- \frac{cos ( 5 x ) csc ^{4} ( 5 x )}{4} + \frac{3}{4} (- \frac{1}{2} cot (5 x) csc (5 x) + \frac{1}{2} ln tan (\frac{5 x}{2}))) : - \frac{1}{4} csc^{4} (5 x) cos (5 x) + \frac{3}{4} (- \frac{1}{2} cot (5 x) csc (5 x) + \frac{1}{2} ln tan (\frac{5 x}{2}))

= - \frac{1}{4} csc^{4} (5 x) cos (5 x) + \frac{3}{4} (- \frac{1}{2} cot (5 x) csc (5 x) + \frac{1}{2} ln tan (\frac{5 x}{2}))

Add a constant to the solution

= - \frac{1}{4} csc^{4} (5 x) cos (5 x) + \frac{3}{4} (- \frac{1}{2} cot (5 x) csc (5 x) + \frac{1}{2} ln tan (\frac{5 x}{2})) + C

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\int \frac{v ^{2}}{( a ^{2} - v ^{2} ) ^{\frac{3}{2}}} d v

\int \frac{e ^{x}}{2 x} d x

What is the integral of 5csc^5(5x) ?
The integral of 5csc^5(5x) is -1/4 csc^4(5x)cos(5x)+3/4 (-1/2 cot(5x)csc(5x)+1/2 ln|tan((5x)/2)|)+C