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

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

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integral of sqrt(2+2cos(θ))

\int 2 + 2 cos (θ) d θ

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

Solution steps

\int 2 + 2 cos (θ) d θ

Apply u-substitution

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

= \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

: \int cos (v) d v

= 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{θ}{2})))

Simplify

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

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

Add a constant to the solution

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

y^{3} - (10 x + 2) + 3 x y^{2} y^{^{'}} = 0

\int 6 e^{7 x} d x

\frac{d}{d x} (\frac{1}{2} x^{3} + \frac{3}{2 x})

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 1, y^{^{'}} (0) = 2

\frac{d}{d x} (c^{1} e^{- x} + c^{2} e^{2 x})

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