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

The integral from 0 to pi of sqrt(1+cos(4x)) is 2sqrt(2)

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integral from 0 to pi of sqrt(1+cos(4x))

\int_{0}^{π} 1 + cos (4 x) d x

22

Decimal

2.82842 \dots

Solution steps

\int_{0}^{π} 1 + cos (4 x) d x

Apply Integral Substitution

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

Rewrite using trig identities

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

Take the constant out:

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

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

Apply exponent rule:

a^{2} = ∣ a ∣

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

Eliminate Absolutes

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

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

\int_{\frac{π}{2}}^{\frac{3 π}{2}} - cos (u) d u = 2

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

= \frac{1}{2} 2 (1 + 2 + 1)

Simplify

\frac{1}{2} 2 (1 + 2 + 1) : 22

= 22

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What is the integral from 0 to pi of sqrt(1+cos(4x)) ?
The integral from 0 to pi of sqrt(1+cos(4x)) is 2sqrt(2)