What is the integral of (1-sin(x))^{1/2} ?

The integral of (1-sin(x))^{1/2} is 2(-cos(x/2)-sin(x/2))+C

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integral of (1-sin(x))^{1/2}

\int (1 - sin (x))^{\frac{1}{2}} d x

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

Solution steps

\int (1 - sin (x))^{\frac{1}{2}} d x

Apply u-substitution

: \int \frac{2 ( u - 1 )}{( 1 + u ^{2} ) ^{\frac{3}{2}}} d u

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

Expand

\frac{u - 1}{( 1 + u ^{2} ) ^{\frac{3}{2}}} : \frac{u}{( 1 + u ^{2} ) ^{\frac{3}{2}}} - \frac{1}{( 1 + u ^{2} ) ^{\frac{3}{2}}}

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

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} ) ^{\frac{3}{2}}} d u - \int \frac{1}{( 1 + u ^{2} ) ^{\frac{3}{2}}} d u)

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

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

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

Substitute back

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

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

Simplify

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

\frac{d}{d x} (ln (2 x))

y^{^{''}} - 5 y^{^{'}} + 4 y = 8 e^{x}

\int_{- 2}^{2} x d x

f (x) = - x^{2} + 16 x + 63, at x = 2

\int ln (4 x) d x

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