What is the integral from-2 to 2 of 3/(4+3t^2) ?

The integral from-2 to 2 of 3/(4+3t^2) is pi/(sqrt(3))

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integral from-2 to 2 of 3/(4+3t^2)

\int_{- 2}^{2} \frac{3}{4 + 3 t ^{2}} d t

\frac{π}{3}

Decimal

1.81379 \dots

Solution steps

\int_{- 2}^{2} \frac{3}{4 + 3 t ^{2}} d t

Take the constant out:

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

= 3 \cdot \int_{- 2}^{2} \frac{1}{4 + 3 t ^{2}} d t

Apply Integral Substitution

= 3 \cdot \int_{- 3}^{3} \frac{1}{2 3 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2 3} \cdot \int_{- 3}^{3} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 3 \cdot \frac{1}{2 3} [arctan (u)]_{- 3}^{3}

Simplify

3 \cdot \frac{1}{2 3} [arctan (u)]_{- 3}^{3} : \frac{3}{2} [arctan (u)]_{- 3}^{3}

= \frac{3}{2} [arctan (u)]_{- 3}^{3}

Compute the boundaries:

2 \cdot \frac{π}{3}

= \frac{3}{2} \cdot 2 \cdot \frac{π}{3}

Simplify

= \frac{π}{3}

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\int_{0}^{1} 4 - x^{2} d x

\int_{0}^{3} x (9 - x^{2}) d x

What is the integral from-2 to 2 of 3/(4+3t^2) ?
The integral from-2 to 2 of 3/(4+3t^2) is pi/(sqrt(3))