What is the integral from 0 to 1 of 5(1+x^2)^{-3/2} ?

The integral from 0 to 1 of 5(1+x^2)^{-3/2} is 5/(sqrt(2))

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integral from 0 to 1 of 5(1+x^2)^{-3/2}

\int_{0}^{1} 5 (1 + x^{2})^{- \frac{3}{2}} d x

\frac{5}{2}

Decimal

3.53553 \dots

Solution steps

\int_{0}^{1} 5 (1 + x^{2})^{- \frac{3}{2}} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{1} (1 + x^{2})^{- \frac{3}{2}} d x

Apply Trigonometric Substitution

= 5 \cdot \int_{0}^{\frac{π}{4}} cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= 5 [sin (u)]_{0}^{\frac{π}{4}}

Compute the boundaries:

\frac{1}{2}

= 5 \cdot \frac{1}{2}

Simplify

= \frac{5}{2}

What is the integral from 0 to 1 of 5(1+x^2)^{-3/2} ?
The integral from 0 to 1 of 5(1+x^2)^{-3/2} is 5/(sqrt(2))