What is the integral of 16sin^2(x) ?

The integral of 16sin^2(x) is 8(x-1/2 sin(2x))+C

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integral of 16sin^2(x)

\int 16 sin^{2} (x) d x

8 (x - \frac{1}{2} sin (2 x)) + C

Solution steps

\int 16 sin^{2} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 16 \cdot \int sin^{2} (x) d x

Rewrite using trig identities

= 16 \cdot \int \frac{1 - cos ( 2 x )}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 16 \cdot \frac{1}{2} \cdot \int 1 - cos (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

= 16 \cdot \frac{1}{2} (\int 1 d x - \int cos (2 x) d x)

\int 1 d x = x

\int cos (2 x) d x = \frac{1}{2} sin (2 x)

= 16 \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x))

Simplify

16 \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x)) : 8 (x - \frac{1}{2} sin (2 x))

= 8 (x - \frac{1}{2} sin (2 x))

Add a constant to the solution

= 8 (x - \frac{1}{2} sin (2 x)) + C

\int 3 x d x

\frac{\partial}{\partial φ} (sin (θ) cos (φ))

\int 1 + 16 x^{2} d x

(x + 1) y^{^{'}} + y = x

x \to - 3 lim (5)

What is the integral of 16sin^2(x) ?
The integral of 16sin^2(x) is 8(x-1/2 sin(2x))+C