What is the integral of cot(x)(-8sin(x)-10csc(x)) ?

The integral of cot(x)(-8sin(x)-10csc(x)) is -8sin(x)+10csc(x)+C

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integral of cot(x)(-8sin(x)-10csc(x))

\int cot (x) (- 8 sin (x) - 10 csc (x)) d x

- 8 sin (x) + 10 csc (x) + C

Solution steps

\int cot (x) (- 8 sin (x) - 10 csc (x)) d x

Rewrite using trig identities

= \int \frac{cos ( x ) ( - 8 sin ^{2} ( x ) - 10 )}{sin ^{2} ( x )} d x

Expand

\frac{cos ( x ) ( - 8 sin ^{2} ( x ) - 10 )}{sin ^{2} ( x )} : - 8 cos (x) - \frac{10 cos ( x )}{sin ^{2} ( x )}

= \int - 8 cos (x) - \frac{10 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 8 cos (x) d x - \int \frac{10 cos ( x )}{sin ^{2} ( x )} d x

\int 8 cos (x) d x = 8 sin (x)

\int \frac{10 cos ( x )}{sin ^{2} ( x )} d x = - 10 csc (x)

= - 8 sin (x) - (- 10 csc (x))

Simplify

= - 8 sin (x) + 10 csc (x)

Add a constant to the solution

= - 8 sin (x) + 10 csc (x) + C

y^{2} + x^{2} y^{^{'}} = 0

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

\int e^{c o s (4 x)} sin (4 x) d x

t^{2} y^{^{''}} + 6 t y^{^{'}} + 6 y = 0

2 x y^{2} + 4 = 2 (3 - x^{2} y) y^{^{'}}

What is the integral of cot(x)(-8sin(x)-10csc(x)) ?
The integral of cot(x)(-8sin(x)-10csc(x)) is -8sin(x)+10csc(x)+C