What is the integral of 8sin^2(x)-2sin(x)-1 ?

The integral of 8sin^2(x)-2sin(x)-1 is 3x-2sin(2x)+2cos(x)+C

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integral of 8sin^2(x)-2sin(x)-1

\int 8 sin^{2} (x) - 2 sin (x) - 1 d x

3 x - 2 sin (2 x) + 2 cos (x) + C

Solution steps

\int 8 sin^{2} (x) - 2 sin (x) - 1 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 sin^{2} (x) d x - \int 2 sin (x) d x - \int 1 d x

\int 8 sin^{2} (x) d x = 4 (x - \frac{1}{2} sin (2 x))

\int 2 sin (x) d x = - 2 cos (x)

\int 1 d x = x

= 4 (x - \frac{1}{2} sin (2 x)) - (- 2 cos (x)) - x

Simplify

4 (x - \frac{1}{2} sin (2 x)) - (- 2 cos (x)) - x : 3 x - 2 sin (2 x) + 2 cos (x)

= 3 x - 2 sin (2 x) + 2 cos (x)

Add a constant to the solution

= 3 x - 2 sin (2 x) + 2 cos (x) + C

f (x) = \frac{3}{x}, at x = 1

f (x) = \frac{3 x ^{2}}{x ^{3} - 8}

4 x^{10 x^{3}}

f (x) = ln (3^{2 - x^{2}})

\int \frac{x + 4}{x + 2} d x

What is the integral of 8sin^2(x)-2sin(x)-1 ?
The integral of 8sin^2(x)-2sin(x)-1 is 3x-2sin(2x)+2cos(x)+C