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

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

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

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

- \frac{1}{3} arcsin (3 cos (x)) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int \frac{1}{3} d v

= - \int \frac{1}{3} d v

Take the constant out:

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

= - \frac{1}{3} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= - \frac{1}{3} \cdot 1 \cdot v

Substitute back

= - \frac{1}{3} \cdot 1 \cdot arcsin (3 cos (x))

Simplify

= - \frac{1}{3} arcsin (3 cos (x))

Add a constant to the solution

= - \frac{1}{3} arcsin (3 cos (x)) + C

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