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

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

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

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

\frac{1}{3} arctan (\frac{9 π}{4})

Decimal

0.47675 \dots

Solution steps

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

Apply Integral Substitution

= \int_{0}^{\frac{9 π}{4}} \frac{1}{3 ( 1 + u ^{2} )} d u

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{9 π}{4}} \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= \frac{1}{3} [arctan (u)]_{0}^{\frac{9 π}{4}}

Compute the boundaries:

arctan (\frac{9 π}{4})

= \frac{1}{3} arctan (\frac{9 π}{4})

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

\int_{0}^{π} (π - x) \cdot cos (n x) d x

\int_{- L}^{0} - x^{2} sin (\frac{nπ x}{L})

\int_{- 8}^{9} (1 - x) d x

\int_{0}^{π} sin^{2} (x) sin (x) d x

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