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

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

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integral of cos^2(θ)-sin^2(θ)

\int cos^{2} (θ) - sin^{2} (θ) d θ

\frac{1}{2} sin (2 θ) + C

Solution steps

\int cos^{2} (θ) - sin^{2} (θ) d θ

Rewrite using trig identities

= \int cos (2 θ) d θ

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 θ

= \frac{1}{2} sin (2 θ)

Add a constant to the solution

= \frac{1}{2} sin (2 θ) + C

x \to 0 lim ((1 - 8 x)^{\frac{1}{x}})

\frac{d y}{d x} = e^{5 y} sin (6 x)

\frac{d}{d x} (\frac{sin ( x )}{tan ( x ) + x})

\frac{1}{( 3 x + 7 )}

\frac{1}{4 + 3 x}

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