What is the integral of (4-3cos(x))/(sin^2(x)) ?

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

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integral of (4-3cos(x))/(sin^2(x))

\int \frac{4 - 3 cos ( x )}{sin ^{2} ( x )} d x

- 4 cot (x) + 3 csc (x) + C

Solution steps

\int \frac{4 - 3 cos ( x )}{sin ^{2} ( x )} d x

Expand

\frac{4 - 3 cos ( x )}{sin ^{2} ( x )} : \frac{4}{sin ^{2} ( x )} - \frac{3 cos ( x )}{sin ^{2} ( x )}

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

\int \frac{4}{sin ^{2} ( x )} d x = - 4 cot (x)

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

= - 4 cot (x) - (- 3 csc (x))

Simplify

= - 4 cot (x) + 3 csc (x)

Add a constant to the solution

= - 4 cot (x) + 3 csc (x) + C

x \to 0 lim (\frac{8 e ^{2 x} - 8 e ^{- 4 x} - 36 x - 6}{3 x ^{2} + 4 x})

\int - x^{n} d x

x^{2} - 8 x - 5

\int \frac{x + 3}{x ^{2} + 16} d x

t \to 8 lim (\frac{t ( t - 5 ) - 24}{t ^{2} - 8 t})

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