What is the integral from 0 to pi/8 of cos^4(x) ?

The integral from 0 to pi/8 of cos^4(x) is (2(2+sqrt(2))^{3/2}sqrt(2-\sqrt{2)}+3(pi/2-1))/(32)

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integral from 0 to pi/8 of cos^4(x)

\int_{0}^{\frac{π}{8}} cos^{4} (x) d x

\frac{2 ( 2 + 2 ) ^{\frac{3}{2}} 2 - 2 + 3 ( \frac{π}{2} - 1 )}{32}

Decimal Notation

0.35528 \dots

Solution steps

\int_{0}^{\frac{π}{8}} cos^{4} (x) d x

Simplify

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

= \int_{0}^{\frac{π}{8}} cos^{3} (x) cos (x) d x

Apply Integration By Parts

: [cos^{3} (x) sin (x) - \int - 3 cos^{2} (x) sin^{2} (x) d x]_{0}^{\frac{π}{8}}

= [cos^{3} (x) sin (x) - \int - 3 cos^{2} (x) sin^{2} (x) d x]_{0}^{\frac{π}{8}}

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

= [cos^{3} (x) sin (x) - (- \frac{3}{8} (x - \frac{1}{4} sin (4 x)))]_{0}^{\frac{π}{8}}

Simplify

[cos^{3} (x) sin (x) - (- \frac{3}{8} (x - \frac{1}{4} sin (4 x)))]_{0}^{\frac{π}{8}} : [\frac{1}{32} (32 cos^{3} (x) sin (x) + 3 (4 x - sin (4 x)))]_{0}^{\frac{π}{8}}

= [\frac{1}{32} (32 cos^{3} (x) sin (x) + 3 (4 x - sin (4 x)))]_{0}^{\frac{π}{8}}

Compute the boundaries:

\frac{2 ( 2 + 2 ) ^{\frac{3}{2}} 2 - 2 + 3 ( \frac{π}{2} - 1 )}{32}

= \frac{2 ( 2 + 2 ) ^{\frac{3}{2}} 2 - 2 + 3 ( \frac{π}{2} - 1 )}{32}

\frac{d}{d x} (x^{2} \cdot x^{- 1})

\int \frac{2 x + 1}{x ^{2} + 2 x + 37} d x

\frac{\partial}{\partial z} (x^{2} yz)

\frac{d}{d x} (sin (x) \cdot sin (x))

\int_{- 9}^{9} 81 - x^{2} d x

What is the integral from 0 to pi/8 of cos^4(x) ?
The integral from 0 to pi/8 of cos^4(x) is (2(2+sqrt(2))^{3/2}sqrt(2-\sqrt{2)}+3(pi/2-1))/(32)