What is the integral of 6/(sin^2(x)) ?

The integral of 6/(sin^2(x)) is -6cot(x)+C

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integral of 6/(sin^2(x))

\int \frac{6}{sin ^{2} ( x )} d x

- 6 cot (x) + C

Solution steps

\int \frac{6}{sin ^{2} ( x )} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{sin ^{2} ( x )} d x

Use the common integral:

\int \frac{1}{sin ^{2} ( x )} d x = - cot (x)

= 6 (- cot (x))

Simplify

= - 6 cot (x)

Add a constant to the solution

= - 6 cot (x) + C

\frac{4 x}{x - 3}, at x = 5

f (x) = \frac{x ^{3}}{3} + x^{2} + 4 x - 10

y = 10 x - x^{2}, (5, 26)

\int \frac{1}{x - ln ^{2} ( x ) + ln ( x ) + 3} d x

\frac{d}{d t} (2 e^{t} + 2 e^{2 t})

What is the integral of 6/(sin^2(x)) ?
The integral of 6/(sin^2(x)) is -6cot(x)+C