What is the integral of 1-(-2sin^2(x+4))^2 ?

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

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integral of 1-(-2sin^2(x+4))^2

\int 1 - (- 2 sin^{2} (x + 4))^{2} d x

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

Solution steps

\int 1 - (- 2 sin^{2} (x + 4))^{2} d x

Apply u-substitution

: \int 1 - 4 sin^{4} (u) d u

= \int 1 - 4 sin^{4} (u) d u

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 u - \int 4 sin^{4} (u) d u

\int 1 d u = u

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

= u - (- sin^{3} (u) cos (u) + \frac{3}{2} u - \frac{3}{4} sin (2 u))

Substitute back

u = x + 4

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

Simplify

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

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

Add a constant to the solution

= - \frac{1}{2} x + sin^{3} (x + 4) cos (x + 4) + \frac{3}{4} sin (2 (x + 4)) - 2 + C

\frac{3}{s}

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2} + 6 x}{x ^{2} - 9})

\int 4096 - x^{2} d x

\int \frac{cos ( 3 π x )}{2} d x

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

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