What is the integral from 0 to 5/8 of 1/((64x^2+1)^{3/2)} ?

The integral from 0 to 5/8 of 1/((64x^2+1)^{3/2)} is 5/(8sqrt(26))

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integral from 0 to 5/8 of 1/((64x^2+1)^{3/2)}

\int_{0}^{\frac{5}{8}} \frac{1}{( 64 x ^{2} + 1 ) ^{\frac{3}{2}}} d x

\frac{5}{8 26}

Decimal

0.12257 \dots

Solution steps

\int_{0}^{\frac{5}{8}} \frac{1}{( 64 x ^{2} + 1 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (5)} \frac{1}{8 sec ( u )} d u

= \int_{0}^{a r c t a n (5)} \frac{1}{8 sec ( u )} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{0}^{a r c t a n (5)} \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{8} \cdot \int_{0}^{a r c t a n (5)} cos (u) d u

Use the common integral:

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

= \frac{1}{8} [sin (u)]_{0}^{a r c t a n (5)}

Compute the boundaries:

\frac{5}{26}

= \frac{1}{8} \cdot \frac{5}{26}

Simplify

= \frac{5}{8 26}

\frac{d y}{d x} = \frac{3 e ^{x}}{y ^{2}}

e^{- t} \cdot (cos (4 t) - 2 sin (4 t))

n = 3 \sum \infty \frac{1}{n ^{2}}

F (x) = 5 x^{5} (x^{3} - 9 x)

x \to \infty lim (\frac{2 x + 1}{2})

What is the integral from 0 to 5/8 of 1/((64x^2+1)^{3/2)} ?
The integral from 0 to 5/8 of 1/((64x^2+1)^{3/2)} is 5/(8sqrt(26))