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

The integral from 0 to pi/4 of cos^4(x) is (3pi+8)/(32)

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

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

\frac{3 π + 8}{32}

Decimal Notation

0.54452 \dots

Solution steps

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

Simplify

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

= \int_{0}^{\frac{π}{4}} 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}{4}}

= [cos^{3} (x) sin (x) - \int - 3 cos^{2} (x) sin^{2} (x) d x]_{0}^{\frac{π}{4}}

\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}{4}}

Simplify

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

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

Compute the boundaries:

\frac{3 π + 8}{32}

= \frac{3 π + 8}{32}

25 x^{4} + 30 x^{3} - 11 x^{2} - 12 x + 4

(5 t + 3 y + 6) d t - d y = 0

y^{^{'}} - 3 y = 9 x

(- 3, - 1), (8, 2)

\frac{\partial}{\partial x} (x^{4} - x^{3} y + x y^{3} - y^{4})

What is the integral from 0 to pi/4 of cos^4(x) ?
The integral from 0 to pi/4 of cos^4(x) is (3pi+8)/(32)