What is the integral from 0 to 2 of 1/(8+2t^2) ?

The integral from 0 to 2 of 1/(8+2t^2) is pi/(16)

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integral from 0 to 2 of 1/(8+2t^2)

\int_{0}^{2} \frac{1}{8 + 2 t ^{2}} d t

\frac{π}{16}

Decimal Notation

0.19634 \dots

Solution steps

\int_{0}^{2} \frac{1}{8 + 2 t ^{2}} d t

Apply Integral Substitution

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

Compute the boundaries:

\frac{π}{4}

= \frac{1}{4} \cdot \frac{π}{4}

Simplify

= \frac{π}{16}

\int (arccos (x)) d x

\frac{d}{d x} (x + x^{2})

\frac{1}{s ^{2}} \cdot \frac{3}{( s + 1 )}

\int \frac{2 x + 1}{( x ^{2} + 4 ) ^{2} ( x + 1 ) ^{3}} d x

\frac{d}{d x} (x^{7} cos (x))

What is the integral from 0 to 2 of 1/(8+2t^2) ?
The integral from 0 to 2 of 1/(8+2t^2) is pi/(16)