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

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

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integral from 0 to pi of sqrt(2+2cos(x))

\int_{0}^{π} 2 + 2 cos (x) d x

4

Solution steps

\int_{0}^{π} 2 + 2 cos (x) d x

Factor

2 + 2 cos (x) : 2 (cos (x) + 1)

= \int_{0}^{π} 2 (cos (x) + 1) d x

Rewrite using trig identities

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

Take the constant out:

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

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

Apply exponent rule:

a^{2} = ∣ a ∣

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

Eliminate Absolutes

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int cos (u) d u = sin (u)

= 2 \cdot 2 [sin (u)]_{0}^{\frac{π}{2}}

Simplify

= 4 [sin (u)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= 4 \cdot 1

Simplify

= 4

\int_{\frac{π}{3}}^{\frac{π}{2}} cos (x) d x

\int_{2}^{4} x^{2} - 2 x d x

\int_{0}^{4} x^{2} + 4 x d x

\int_{- 2}^{3} \frac{3}{4 x ^{2}} d x

\int_{0}^{2 π} x cos (n x) d x

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