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

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

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

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

\frac{3 π}{16}

Solution steps

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

Simplify

= \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_{0}^{t} 12 e^{2 x} - 12 e^{x} d x

\int_{0}^{1} \frac{e ^{z} + 3}{e ^{z} + 3 z} d z

\int_{3}^{4} t^{2} - 2 t - 3 d t

\int_{0}^{1} 3^{x} d x

\int_{- π}^{0} - sin (x) d x

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