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

The integral from 0 to pi of (cos(2x))^4 is (3pi)/8

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

\int_{0}^{π} (cos (2 x))^{4} d x

\frac{3 π}{8}

Decimal Notation

1.17809 \dots

Solution steps

\int_{0}^{π} (cos (2 x))^{4} d x

Simplify

= \int_{0}^{π} cos^{4} (2 x) d x

Apply u-substitution

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

= \int_{0}^{2 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}^{2 π} cos^{4} (u) d u

Simplify

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

= \frac{1}{2} \cdot \int_{0}^{2 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}^{2 π}

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

Simplify

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

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

Compute the boundaries:

\frac{3 π}{4}

= \frac{1}{2} \cdot \frac{3 π}{4}

Simplify

= \frac{3 π}{8}

\int 3 x \cdot e^{(- 4 x)} d x

\frac{d}{d x} (\frac{x ^{2} y}{x ^{2} + y ^{2}})

y = x^{2} - 8

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 x^{2}

\int \frac{x}{2 x} d x

What is the integral from 0 to pi of (cos(2x))^4 ?
The integral from 0 to pi of (cos(2x))^4 is (3pi)/8