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

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

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integral of 1-sin^2(x)

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

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

Solution steps

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

Rewrite using trig identities

= \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

= \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

= \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)

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

Add a constant to the solution

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

\frac{d}{d x} ((x - 4)^{2})

\frac{d}{d x} (ln (2 x - x))

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

\int (\frac{5}{x} + \frac{15}{x ^{6}}) d x

t - 4

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