What is the integral from 0 to 4/3 of 1/(16+9x^2) ?

The integral from 0 to 4/3 of 1/(16+9x^2) is pi/(48)

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integral from 0 to 4/3 of 1/(16+9x^2)

\int_{0}^{\frac{4}{3}} \frac{1}{16 + 9 x ^{2}} d x

\frac{π}{48}

Decimal Notation

0.06544 \dots

Solution steps

\int_{0}^{\frac{4}{3}} \frac{1}{16 + 9 x ^{2}} d x

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{12 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{12} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{12} \cdot \frac{π}{4}

Simplify

= \frac{π}{48}

\int sin (9 x) cos (4 x) d x

\int_{1}^{2} I n x d x

y^{^{''}} + 2 y^{^{'}} + 6 y = 0

y^{^{''}} + 25 y = 20 sin (5 t)

2 s

What is the integral from 0 to 4/3 of 1/(16+9x^2) ?
The integral from 0 to 4/3 of 1/(16+9x^2) is pi/(48)