What is the integral of sqrt(2+2cos(x)) ?

The integral of sqrt(2+2cos(x)) is 4tan(x/2)cos(x/2)+C

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integral of sqrt(2+2cos(x))

\int 2 + 2 cos (x) d x

4 tan (\frac{x}{2}) cos (\frac{x}{2}) + C

Solution steps

\int 2 + 2 cos (x) d x

Apply u-substitution

= \int \frac{4}{( 1 + u ^{2} ) 1 + u ^{2}} d u

Take the constant out:

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

= 4 \cdot \int \frac{1}{( 1 + u ^{2} ) 1 + u ^{2}} d u

Simplify

(1 + u^{2}) 1 + u^{2} : (1 + u^{2})^{\frac{3}{2}}

= 4 \cdot \int \frac{1}{( 1 + u ^{2} ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

= 4 \cdot \int cos (v) d v

Use the common integral:

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

= 4 sin (v)

Substitute back

= 4 sin (arctan (tan (\frac{x}{2})))

Simplify

4 sin (arctan (tan (\frac{x}{2}))) : 4 tan (\frac{x}{2}) cos (\frac{x}{2})

= 4 tan (\frac{x}{2}) cos (\frac{x}{2})

Add a constant to the solution

= 4 tan (\frac{x}{2}) cos (\frac{x}{2}) + C

y = 2 x^{2}, y = 8 x^{2}, 5 x + y = 3

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

\int_{7}^{9} \frac{7}{( x - 8 ) ^{10}} d x

n = 1 \sum \infty \frac{1}{2} (0.3)^{n}

\int \frac{1 - x ^{2}}{1 - 2 x} d x

What is the integral of sqrt(2+2cos(x)) ?
The integral of sqrt(2+2cos(x)) is 4tan(x/2)cos(x/2)+C