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

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

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integral of sqrt(1+sin(x))

\int 1 + sin (x) d x

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

Solution steps

\int 1 + sin (x) d x

Apply u-substitution

= \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

d er i v a t i v e f (x) = 5 + 4 e^{3 x}

y^{^{''}} + 16 y^{^{'}} + 64 y = 0, y (0) = - 3, y^{^{'}} (0) = 28

\int_{0}^{8} 1 + \frac{81}{4} x d x

d er i v a t i v e 4 x^{2} - 6 x + 2

\frac{d}{d x} (5 sin (x) (\frac{4}{x ^{3}} + 3 x))

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