What is the integral from 0 to pi/4 of 1 (cos^4(x)) ?

The integral from 0 to pi/4 of 1 (cos^4(x)) is ((3pi+8)d)/(32)

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integral from 0 to pi/4 of 1 (cos^4(x))

\int_{0}^{\frac{π}{4}} \frac{d}{1} (cos^{4} (x)) d x

\frac{( 3 π + 8 ) d}{32}

Solution steps

\int_{0}^{\frac{π}{4}} \frac{d}{1} cos^{4} (x) d x

Take the constant out:

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

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

Simplify

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

= d \cdot \int_{0}^{\frac{π}{4}} cos^{3} (x) cos (x) d x

Apply Integration By Parts

= d [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))

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

Simplify

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

= d [\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}

= d \frac{3 π + 8}{32}

Simplify

= \frac{( 3 π + 8 ) d}{32}

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What is the integral from 0 to pi/4 of 1 (cos^4(x)) ?
The integral from 0 to pi/4 of 1 (cos^4(x)) is ((3pi+8)d)/(32)