What is the integral from 0 to sin(t) of 1/(9-u^2) ?

The integral from 0 to sin(t) of 1/(9-u^2) is (ln| 1/3 sin(t)+1|-ln| 1/3 sin(t)-1|)/6

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integral from 0 to sin(t) of 1/(9-u^2)

\int_{0}^{s i n (t)} \frac{1}{9 - u ^{2}} d u

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

Solution steps

\int_{0}^{s i n (t)} \frac{1}{9 - u ^{2}} d u

Apply Integral Substitution

= \int_{0}^{\frac{s i n ( t )}{3}} \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_{0}^{\frac{s i n ( t )}{3}} \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}]_{0}^{\frac{s i n ( t )}{3}}

Simplify

\frac{1}{3} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{\frac{s i n ( t )}{3}} : \frac{1}{3} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{\frac{s i n ( t )}{3}}

= \frac{1}{3} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{\frac{s i n ( t )}{3}}

Compute the boundaries:

\frac{ln \frac{1}{3} sin ( t ) + 1 - ln \frac{1}{3} sin ( t ) - 1}{2}

= \frac{1}{3} \cdot \frac{ln \frac{1}{3} sin ( t ) + 1 - ln \frac{1}{3} sin ( t ) - 1}{2}

Simplify

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

\int_{0}^{x} y d y

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What is the integral from 0 to sin(t) of 1/(9-u^2) ?
The integral from 0 to sin(t) of 1/(9-u^2) is (ln| 1/3 sin(t)+1|-ln| 1/3 sin(t)-1|)/6