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

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

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

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

\frac{3 π}{16}

Decimal Notation

0.58904 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{π} cos^{4} (u) \frac{1}{2} d u

= \int_{0}^{π} cos^{4} (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{π} cos^{4} (u) d u

Simplify

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

= \frac{1}{2} \cdot \int_{0}^{π} cos^{3} (u) cos (u) d u

Apply Integration By Parts

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

= \frac{1}{2} [cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u]_{0}^{π}

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

= \frac{1}{2} [cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u)))]_{0}^{π}

Simplify

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

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

Compute the boundaries:

\frac{3 π}{8}

= \frac{1}{2} \cdot \frac{3 π}{8}

Simplify

= \frac{3 π}{16}

x \frac{d y}{d x} - y = 2 x^{2} y^{2}

sin (2) x^{3}

x \to 2 + lim (x - 2)

f (x) = arccsc (x)

f (x) = ln (tan (2 x)), at x = \frac{π}{8}

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