What is the integral of 3cos(x)sqrt(1-9sin^2(x)) ?

The integral of 3cos(x)sqrt(1-9sin^2(x)) is 1/2 (arcsin(3sin(x))+1/2 sin(2arcsin(3sin(x))))+C

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integral of 3cos(x)sqrt(1-9sin^2(x))

\int 3 cos (x) 1 - 9 sin^{2} (x) d x

\frac{1}{2} (arcsin (3 sin (x)) + \frac{1}{2} sin (2 arcsin (3 sin (x)))) + C

Solution steps

\int 3 cos (x) 1 - 9 sin^{2} (x) d x

Take the constant out:

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

= 3 \cdot \int cos (x) 1 - 9 sin^{2} (x) d x

Apply u-substitution

: \int 1 - 9 u^{2} d u

= 3 \cdot \int 1 - 9 u^{2} d u

Apply Trigonometric Substitution

: \int \frac{1}{3} cos^{2} (v) d v

= 3 \cdot \int \frac{1}{3} cos^{2} (v) d v

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int cos^{2} (v) d v

Rewrite using trig identities

= 3 \cdot \frac{1}{3} \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 \cdot \frac{1}{3} \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

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

= 3 \cdot \frac{1}{3} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 3 \cdot \frac{1}{3} \cdot \frac{1}{2} (arcsin (3 sin (x)) + \frac{1}{2} sin (2 arcsin (3 sin (x))))

Simplify

3 \cdot \frac{1}{3} \cdot \frac{1}{2} (arcsin (3 sin (x)) + \frac{1}{2} sin (2 arcsin (3 sin (x)))) : \frac{1}{2} (arcsin (3 sin (x)) + \frac{1}{2} sin (2 arcsin (3 sin (x))))

= \frac{1}{2} (arcsin (3 sin (x)) + \frac{1}{2} sin (2 arcsin (3 sin (x))))

Add a constant to the solution

= \frac{1}{2} (arcsin (3 sin (x)) + \frac{1}{2} sin (2 arcsin (3 sin (x)))) + C

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What is the integral of 3cos(x)sqrt(1-9sin^2(x)) ?
The integral of 3cos(x)sqrt(1-9sin^2(x)) is 1/2 (arcsin(3sin(x))+1/2 sin(2arcsin(3sin(x))))+C