What is the integral of 6cos(x)-9/(sqrt(1-x^2)) ?

The integral of 6cos(x)-9/(sqrt(1-x^2)) is 6sin(x)-9arcsin(x)+C

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integral of 6cos(x)-9/(sqrt(1-x^2))

\int 6 cos (x) - \frac{9}{1 - x ^{2}} d x

6 sin (x) - 9 arcsin (x) + C

Solution steps

\int 6 cos (x) - \frac{9}{1 - x ^{2}} 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 6 cos (x) d x - \int \frac{9}{1 - x ^{2}} d x

\int 6 cos (x) d x = 6 sin (x)

\int \frac{9}{1 - x ^{2}} d x = 9 arcsin (x)

= 6 sin (x) - 9 arcsin (x)

Add a constant to the solution

= 6 sin (x) - 9 arcsin (x) + C

\int - 2 e^{- \frac{1}{2} x} d x

3 sin (5 t + 4 5^{\circ})

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

\frac{d}{d x} (\frac{1}{4 - x})

\int \frac{1}{3 x - 7} d x

What is the integral of 6cos(x)-9/(sqrt(1-x^2)) ?
The integral of 6cos(x)-9/(sqrt(1-x^2)) is 6sin(x)-9arcsin(x)+C