What is the integral of sqrt(36-36cos^2(x)) ?

The integral of sqrt(36-36cos^2(x)) is -6cos(x)+C

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integral of sqrt(36-36cos^2(x))

\int 36 - 36 cos^{2} (x) d x

- 6 cos (x) + C

Solution steps

\int 36 - 36 cos^{2} (x) d x

Rewrite using trig identities

= \int 6 sin^{2} (x) d x

Take the constant out:

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

= 6 \cdot \int sin^{2} (x) d x

sin^{2} (x) = (sin (x)),

assuming

sin (x) \geq 0

= 6 \cdot \int sin (x) d x

Use the common integral:

\int sin (x) d x = - cos (x)

= 6 (- cos (x))

Simplify

= - 6 cos (x)

Add a constant to the solution

= - 6 cos (x) + C

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What is the integral of sqrt(36-36cos^2(x)) ?
The integral of sqrt(36-36cos^2(x)) is -6cos(x)+C