What is the integral from (sqrt(2))/4 to 1/2 of 1/(x^5sqrt(16x^2-1)) ?

The integral from (sqrt(2))/4 to 1/2 of 1/(x^5sqrt(16x^2-1)) is 4(2pi+7sqrt(3)-16)

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integral from (sqrt(2))/4 to 1/2 of 1/(x^5sqrt(16x^2-1))

\int_{\frac{2}{4}}^{\frac{1}{2}} \frac{1}{x ^{5} 16 x ^{2} - 1} d x

4 (2 π + 73 - 16)

Decimal Notation

9.63016 \dots

Solution steps

\int_{\frac{2}{4}}^{\frac{1}{2}} \frac{1}{x ^{5} 16 x ^{2} - 1} d x

Apply Trigonometric Substitution

: \int_{\frac{π}{4}}^{\frac{π}{3}} 256 cos^{4} (u) d u

= \int_{\frac{π}{4}}^{\frac{π}{3}} 256 cos^{4} (u) d u

Take the constant out:

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

= 256 \cdot \int_{\frac{π}{4}}^{\frac{π}{3}} cos^{4} (u) d u

Simplify

cos^{4} (u) : cos^{3} (u) cos (u)

= 256 \cdot \int_{\frac{π}{4}}^{\frac{π}{3}} cos^{3} (u) cos (u) d u

Apply Integration By Parts

: [cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u]_{\frac{π}{4}}^{\frac{π}{3}}

= 256 [cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u]_{\frac{π}{4}}^{\frac{π}{3}}

\int - 3 cos^{2} (u) sin^{2} (u) d u = - \frac{3}{8} (u - \frac{1}{4} sin (4 u))

= 256 [cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u)))]_{\frac{π}{4}}^{\frac{π}{3}}

Simplify

256 [cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u)))]_{\frac{π}{4}}^{\frac{π}{3}} : 256 [\frac{1}{32} (32 cos^{3} (u) sin (u) + 3 (4 u - sin (4 u)))]_{\frac{π}{4}}^{\frac{π}{3}}

= 256 [\frac{1}{32} (32 cos^{3} (u) sin (u) + 3 (4 u - sin (4 u)))]_{\frac{π}{4}}^{\frac{π}{3}}

Compute the boundaries:

\frac{2 π + 7 3 - 16}{64}

= 256 \cdot \frac{2 π + 7 3 - 16}{64}

Simplify

= 4 (2 π + 73 - 16)

x \to 0 - lim (x^{2} \cdot sin (\frac{1}{x}))

\int_{- 2}^{3} \frac{33}{3 - x} d x

\int - 3 cos^{2} (x) sin^{2} (x) d x

\int \frac{7 x ^{2}}{( 36 + x ^{2} ) ^{2}} d x

\frac{d y}{d x} = \frac{2 x - 1}{sin ( y )}

What is the integral from (sqrt(2))/4 to 1/2 of 1/(x^5sqrt(16x^2-1)) ?
The integral from (sqrt(2))/4 to 1/2 of 1/(x^5sqrt(16x^2-1)) is 4(2pi+7sqrt(3)-16)