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)

integral

1 - sin^{2} (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

1 - ∣ x ∣

(1 - x)^{3}

e^{2} x^{2}

\frac{1}{( 2 x + 1 ) ^{3}}

\frac{cos ( \frac{1}{x} )}{x ^{3}}

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