What is the integral of 1/(9-10sin^2(x)) ?

The integral of 1/(9-10sin^2(x)) is 1/6 (ln|(tan(x))/3+1|-ln|(tan(x))/3-1|)+C

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integral of 1/(9-10sin^2(x))

\int \frac{1}{9 - 10 sin ^{2} ( x )} d x

\frac{1}{6} (ln \frac{tan ( x )}{3} + 1 - ln \frac{tan ( x )}{3} - 1) + C

Solution steps

\int \frac{1}{9 - 10 sin ^{2} ( x )} d x

Apply Trigonometric 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{t a n ( x )}{3} + 1}{2} - \frac{ln \frac{t a n ( x )}{3} - 1}{2}

Simplify

\frac{1}{3} \frac{ln \frac{t a n ( x )}{3} + 1}{2} - \frac{ln \frac{t a n ( x )}{3} - 1}{2} : \frac{1}{6} (ln \frac{tan ( x )}{3} + 1 - ln \frac{tan ( x )}{3} - 1)

= \frac{1}{6} (ln \frac{tan ( x )}{3} + 1 - ln \frac{tan ( x )}{3} - 1)

Add a constant to the solution

= \frac{1}{6} (ln \frac{tan ( x )}{3} + 1 - ln \frac{tan ( x )}{3} - 1) + C

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What is the integral of 1/(9-10sin^2(x)) ?
The integral of 1/(9-10sin^2(x)) is 1/6 (ln|(tan(x))/3+1|-ln|(tan(x))/3-1|)+C