What is the integral from 0 to pi/2 of cos^4(x) ?

The integral from 0 to pi/2 of cos^4(x) is (3pi)/(16)

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integral from 0 to pi/2 of cos^4(x)

\int_{0}^{\frac{π}{2}} cos^{4} (x) d x

\frac{3 π}{16}

Decimal

0.58904 \dots

Solution steps

\int_{0}^{\frac{π}{2}} cos^{4} (x) d x

Simplify

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

= \int_{0}^{\frac{π}{2}} 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}{2}}

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

\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}{2}}

Simplify

[cos^{3} (x) sin (x) - (- \frac{3}{8} (x - \frac{1}{4} sin (4 x)))]_{0}^{\frac{π}{2}} : [\frac{1}{32} (32 cos^{3} (x) sin (x) + 3 (4 x - sin (4 x)))]_{0}^{\frac{π}{2}}

= [\frac{1}{32} (32 cos^{3} (x) sin (x) + 3 (4 x - sin (4 x)))]_{0}^{\frac{π}{2}}

Compute the boundaries:

\frac{3 π}{16}

= \frac{3 π}{16}

\int \frac{5 x + 3}{x ^{2} + 4 x + 4} d x

\frac{d}{d x} (2^{3^{x^{2}}})

\int x^{2} x^{3} + 29 d x

(1 - x^{2}) y^{^{'}} + x (y - 1) = - x^{3}

(3 x y + y^{2}) + (x^{2} + x y) y^{^{'}} = 0

What is the integral from 0 to pi/2 of cos^4(x) ?
The integral from 0 to pi/2 of cos^4(x) is (3pi)/(16)