What is the integral of 1-(16)/(sin^4(x)) ?

The integral of 1-(16)/(sin^4(x)) is x-16(-1/3 csc^3(x)cos(x)-2/3 cot(x))+C

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integral of 1-(16)/(sin^4(x))

\int 1 - \frac{16}{sin ^{4} ( x )} d x

x - 16 (- \frac{1}{3} csc^{3} (x) cos (x) - \frac{2}{3} cot (x)) + C

Solution steps

\int 1 - \frac{16}{sin ^{4} ( x )} d x

Apply the Sum Rule:

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

= \int 1 d x - \int \frac{16}{sin ^{4} ( x )} d x

\int 1 d x = x

\int \frac{16}{sin ^{4} ( x )} d x = 16 (- \frac{1}{3} csc^{3} (x) cos (x) - \frac{2}{3} cot (x))

= x - 16 (- \frac{1}{3} csc^{3} (x) cos (x) - \frac{2}{3} cot (x))

Add a constant to the solution

= x - 16 (- \frac{1}{3} csc^{3} (x) cos (x) - \frac{2}{3} cot (x)) + C

x \to 0 lim (x ln^{2} (x))

\frac{\partial}{\partial x} (3 y^{2} - x^{2})

\frac{d}{d x} (\frac{x ^{2} + 4}{4 x ^{3} - 5 x})

- 5 e^{- 5 x + 2}

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

What is the integral of 1-(16)/(sin^4(x)) ?
The integral of 1-(16)/(sin^4(x)) is x-16(-1/3 csc^3(x)cos(x)-2/3 cot(x))+C