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

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

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integral of cos^2(t)-sin^2(t)

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

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

Solution steps

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

Rewrite using trig identities

= \int cos (2 t) d t

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 t

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (sin (x^{1} y^{3}))

(- 2 cos (2 x) - sin (3 x))^{2}

x \to \frac{1}{2} lim (arcsin (x))

f (x) = x^{2} + x - 3

3 t^{4} y^{^{''}} + 7 t^{3} y^{^{'}} = 0

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