What is the integral from 0 to 1/10 of 1/((100x^2+1)^{3/2)} ?

The integral from 0 to 1/10 of 1/((100x^2+1)^{3/2)} is 1/(10sqrt(2))

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integral from 0 to 1/10 of 1/((100x^2+1)^{3/2)}

\int_{0}^{\frac{1}{10}} \frac{1}{( 100 x ^{2} + 1 ) ^{\frac{3}{2}}} d x

\frac{1}{10 2}

Decimal Notation

0.07071 \dots

Solution steps

\int_{0}^{\frac{1}{10}} \frac{1}{( 100 x ^{2} + 1 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} \frac{1}{10 sec ( u )} d u

= \int_{0}^{\frac{π}{4}} \frac{1}{10 sec ( u )} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int_{0}^{\frac{π}{4}} \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{10} \cdot \int_{0}^{\frac{π}{4}} cos (u) d u

Use the common integral:

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

= \frac{1}{10} [sin (u)]_{0}^{\frac{π}{4}}

Compute the boundaries:

\frac{1}{2}

= \frac{1}{10} \cdot \frac{1}{2}

Simplify

= \frac{1}{10 2}

\int \frac{( sin ( x ) )}{cos ^{2} ( x )} d x

\frac{d y}{d x}, 9 x y = 5

\frac{s + 5}{( s + 2 ) ( s + 3 )}

x \to \infty lim (\frac{8}{x} - 8)

y = \frac{cos ( x )}{2 + sin ( x )}

What is the integral from 0 to 1/10 of 1/((100x^2+1)^{3/2)} ?
The integral from 0 to 1/10 of 1/((100x^2+1)^{3/2)} is 1/(10sqrt(2))