What is the integral from 0 to 2/7 of 1/(49x^2+4) ?

The integral from 0 to 2/7 of 1/(49x^2+4) is pi/(56)

Popular >

integral from 0 to 2/7 of 1/(49x^2+4)

\int_{0}^{\frac{2}{7}} \frac{1}{49 x ^{2} + 4} d x

\frac{π}{56}

Decimal Notation

0.05609 \dots

Solution steps

\int_{0}^{\frac{2}{7}} \frac{1}{49 x ^{2} + 4} d x

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{14 ( 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}{14} \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}{14} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{14} \cdot \frac{π}{4}

Simplify

= \frac{π}{56}

\int_{1}^{3} e^{- x} d x

\int_{0}^{1} x y x^{2} + y^{2} d y

\int_{0}^{2 π} - sin^{2} (θ) d θ

\int_{- 2}^{1} (3 x + 4) d x

\int_{- 1}^{1} - x^{3} d x

What is the integral from 0 to 2/7 of 1/(49x^2+4) ?
The integral from 0 to 2/7 of 1/(49x^2+4) is pi/(56)