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

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

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integral of-sin^2(x)+cos^2(x)

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

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

Solution steps

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

Rewrite using trig identities

= \int cos (2 x) d x

Apply u-substitution

: \int cos (u) \frac{1}{2} d u

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

Take the constant out:

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

= \frac{1}{2} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{2} sin (u)

Substitute back

u = 2 x

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

Add a constant to the solution

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

\int_{1}^{3} \frac{x ^{2} - 1}{x} d x

\int arctan (x - 1) d x

\int \frac{1}{( x ^{3} - 1 )} d x

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

\frac{d}{d x} (tan (1 - x))

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