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

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

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integral of sqrt(2cos(θ)+2)

\int 2 cos (θ) + 2 d θ

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

Solution steps

\int 2 cos (θ) + 2 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

\int \frac{1}{49 e ^{- 8 x} + e ^{8 x}} d x

\frac{d}{d x} (2 sin^{2} (π x))

\frac{\partial}{\partial x} (ln (x - 3 y))

\frac{d}{d x} (1 + cos (3 π x))

\int \frac{1}{1 + cos ( u )} d u

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