What is the integral of sqrt(cos(x)+1) ?

The integral of sqrt(cos(x)+1) is 2sqrt(2)tan(x/2)cos(x/2)+C

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integral of sqrt(cos(x)+1)

\int cos (x) + 1 d x

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

Solution steps

\int cos (x) + 1 d x

Apply u-substitution

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

= \int \frac{2 2}{( 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

= 22 \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}}

= 22 \cdot \int \frac{1}{( 1 + u ^{2} ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

: \int cos (v) d v

= 22 \cdot \int cos (v) d v

Use the common integral:

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

= 22 sin (v)

Substitute back

= 22 sin (arctan (tan (\frac{x}{2})))

Simplify

22 sin (arctan (tan (\frac{x}{2}))) : 22 tan (\frac{x}{2}) cos (\frac{x}{2})

= 22 tan (\frac{x}{2}) cos (\frac{x}{2})

Add a constant to the solution

= 22 tan (\frac{x}{2}) cos (\frac{x}{2}) + C

\int (4 - x) d x

\frac{d}{d y} (y^{2} - 9)

\frac{d}{d x} (2^{4 - x^{2}})

\int ln (x) + 2 d x

7 y^{^{'}} - 21 x^{2} = 0

What is the integral of sqrt(cos(x)+1) ?
The integral of sqrt(cos(x)+1) is 2sqrt(2)tan(x/2)cos(x/2)+C