What is the integral from 0 to x of sqrt(1-sin(x)) ?

The integral from 0 to x of sqrt(1-sin(x)) is 2sqrt(sin(x)+1)-2

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integral from 0 to x of sqrt(1-sin(x))

\int_{0}^{x} 1 - sin (x) d x

2 sin (x) + 1 - 2

Solution steps

\int_{0}^{x} 1 - sin (x) d x

Apply u-substitution

: \int_{0}^{s i n (x)} \frac{1}{u + 1} d u

= \int_{0}^{s i n (x)} \frac{1}{u + 1} d u

Apply u-substitution

: \int_{1}^{s i n (x) + 1} \frac{1}{v} d v

= \int_{1}^{s i n (x) + 1} \frac{1}{v} d v

Simplify

\frac{1}{v} : \frac{1}{v}

= \int_{1}^{s i n (x) + 1} \frac{1}{v} d v

Simplify

= \int_{1}^{s i n (x) + 1} \frac{1}{v} d v

Apply the Power Rule

: [2 v^{\frac{1}{2}}]_{1}^{s i n (x) + 1}

= [2 v^{\frac{1}{2}}]_{1}^{s i n (x) + 1}

Apply radical rule:

a^{\frac{1}{n}} = n a

= [2 v]_{1}^{s i n (x) + 1}

Compute the boundaries:

2 sin (x) + 1 - 2

= 2 sin (x) + 1 - 2

\int_{0}^{\frac{1}{2}} 3 x d x

\int_{\frac{1}{2}}^{2} (5 - \frac{5}{x ^{2}}) d x

\int_{1}^{2} x^{- 1} + x^{2} d x

\int_{1}^{32} x^{\frac{- 6}{5}} d x

\int_{1}^{\infty} x \cdot \frac{3}{x ^{4}} d x

What is the integral from 0 to x of sqrt(1-sin(x)) ?
The integral from 0 to x of sqrt(1-sin(x)) is 2sqrt(sin(x)+1)-2