What is the integral of (cos(θ))/(9-sin^2(θ)) ?

The integral of (cos(θ))/(9-sin^2(θ)) is 1/6 (ln|(sin(θ))/3+1|-ln|(sin(θ))/3-1|)+C

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integral of (cos(θ))/(9-sin^2(θ))

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

\frac{1}{6} (ln \frac{sin ( θ )}{3} + 1 - ln \frac{sin ( θ )}{3} - 1) + C

Solution steps

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

Apply u-substitution

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

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

Apply Integral Substitution

= \int \frac{1}{3 ( - v ^{2} + 1 )} 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 \frac{1}{- v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{3} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= \frac{1}{3} \frac{ln \frac{s i n ( θ )}{3} + 1}{2} - \frac{ln \frac{s i n ( θ )}{3} - 1}{2}

Simplify

\frac{1}{3} \frac{ln \frac{s i n ( θ )}{3} + 1}{2} - \frac{ln \frac{s i n ( θ )}{3} - 1}{2} : \frac{1}{6} (ln \frac{sin ( θ )}{3} + 1 - ln \frac{sin ( θ )}{3} - 1)

= \frac{1}{6} (ln \frac{sin ( θ )}{3} + 1 - ln \frac{sin ( θ )}{3} - 1)

Add a constant to the solution

= \frac{1}{6} (ln \frac{sin ( θ )}{3} + 1 - ln \frac{sin ( θ )}{3} - 1) + C

\frac{d}{d x} (\frac{y}{x ^{2} + y ^{2}})

f (x) = 5 x^{3} - 5 x, at x = - 1

\frac{d}{d x} (cos^{n - 1} (x))

y = x^{4} - 4 x^{3} + 10

2 x, x ∣ x + 2 ∣

What is the integral of (cos(θ))/(9-sin^2(θ)) ?
The integral of (cos(θ))/(9-sin^2(θ)) is 1/6 (ln|(sin(θ))/3+1|-ln|(sin(θ))/3-1|)+C