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

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

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integral of sqrt(sin(x)+1)

\int sin (x) + 1 d x

2 (- cos (\frac{x}{2}) + sin (\frac{x}{2})) + C

Solution steps

\int sin (x) + 1 d x

Apply u-substitution

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

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

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= 2 (- \frac{1}{1 + u ^{2}} + \frac{u}{1 + u ^{2}})

Substitute back

u = tan (\frac{x}{2})

= 2 - \frac{1}{1 + tan ^{2} ( \frac{x}{2} )} + \frac{tan ( \frac{x}{2} )}{1 + tan ^{2} ( \frac{x}{2} )}

Simplify

2 - \frac{1}{1 + tan ^{2} ( \frac{x}{2} )} + \frac{tan ( \frac{x}{2} )}{1 + tan ^{2} ( \frac{x}{2} )} : 2 (- cos (\frac{x}{2}) + sin (\frac{x}{2}))

= 2 (- cos (\frac{x}{2}) + sin (\frac{x}{2}))

Add a constant to the solution

= 2 (- cos (\frac{x}{2}) + sin (\frac{x}{2})) + C

\int \frac{1}{( 4 x ^{2} + 9 ) ^{\frac{3}{2}}} d x

n = 0 \sum \infty 1 + 2 + 3 + 4 + 5

\frac{d}{d x} (\frac{e ^{a r c t a n (x)}}{x ^{2} + 1})

\int (\frac{x}{2} + 1) d x

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

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