What is the integral of csc^2(0.1x) ?

The integral of csc^2(0.1x) is -10cot(0.1x)+C

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integral of csc^2(0.1x)

\int csc^{2} (0.1 x) d x

- 10 cot (0.1 x) + C

Solution steps

\int csc^{2} (0.1 x) d x

Apply u-substitution

: \int csc^{2} (u) \frac{1}{0.1} d u

= \int csc^{2} (u) \frac{1}{0.1} d u

Take the constant out:

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

= \frac{1}{0.1} \cdot \int csc^{2} (u) d u

Simplify

= 10 \cdot \int csc^{2} (u) d u

Use the common integral:

\int csc^{2} (u) d u = - cot (u)

= 10 (- cot (u))

Substitute back

u = 0.1 x

= 10 (- cot (0.1 x))

Simplify

= - 10 cot (0.1 x)

Add a constant to the solution

= - 10 cot (0.1 x) + C

Q

x \to 0 lim (\frac{ln ( 1 + 4 x )}{x})

\int x^{11} x^{6} + 5 d x

ln (x - 6)

x^{2} y^{^{''}} - 20 y = 0

What is the integral of csc^2(0.1x) ?
The integral of csc^2(0.1x) is -10cot(0.1x)+C