What is the integral from 0 to 4 of 1/(36x^2+49) ?

The integral from 0 to 4 of 1/(36x^2+49) is 1/42 arctan(24/7)

Popular >

integral from 0 to 4 of 1/(36x^2+49)

\int_{0}^{4} \frac{1}{36 x ^{2} + 49} d x

\frac{1}{42} arctan (\frac{24}{7})

Decimal

0.03064 \dots

Solution steps

\int_{0}^{4} \frac{1}{36 x ^{2} + 49} d x

Apply Integral Substitution

= \int_{0}^{\frac{24}{7}} \frac{1}{42 ( 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}{42} \cdot \int_{0}^{\frac{24}{7}} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{42} [arctan (u)]_{0}^{\frac{24}{7}}

Compute the boundaries:

arctan (\frac{24}{7})

= \frac{1}{42} arctan (\frac{24}{7})

\int_{0}^{2 π} cos (\frac{x}{2}) d x

\int_{- 2}^{2} \frac{1}{1 + x} d x

\int_{0}^{3 x} cos (t) d t

\int_{0}^{\infty} x^{7} e^{- x} d x

\int_{0}^{1} 6 t 1 + t^{2} d t

What is the integral from 0 to 4 of 1/(36x^2+49) ?
The integral from 0 to 4 of 1/(36x^2+49) is 1/42 arctan(24/7)