What is the integral from-1 to 1 of 6/(1+36t^2) ?

The integral from-1 to 1 of 6/(1+36t^2) is 2arctan(6)

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integral from-1 to 1 of 6/(1+36t^2)

\int_{- 1}^{1} \frac{6}{1 + 36 t ^{2}} d t

2 arctan (6)

Decimal Notation

2.81129 \dots

Solution steps

\int_{- 1}^{1} \frac{6}{1 + 36 t ^{2}} d t

Take the constant out:

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

= 6 \cdot \int_{- 1}^{1} \frac{1}{1 + 36 t ^{2}} d t

Apply Integral Substitution

= 6 \cdot \int_{- 6}^{6} \frac{1}{6 ( 1 + u ^{2} )} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int_{- 6}^{6} \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= 6 \cdot \frac{1}{6} [arctan (u)]_{- 6}^{6}

Simplify

6 \cdot \frac{1}{6} [arctan (u)]_{- 6}^{6} : [arctan (u)]_{- 6}^{6}

= [arctan (u)]_{- 6}^{6}

Compute the boundaries:

2 arctan (6)

= 2 arctan (6)

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What is the integral from-1 to 1 of 6/(1+36t^2) ?
The integral from-1 to 1 of 6/(1+36t^2) is 2arctan(6)