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

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

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integral from 0 to sin(x) of sqrt(1-t^2)

\int_{0}^{s i n (x)} 1 - t^{2} d t

\frac{1}{2} (arcsin (sin (x)) + \frac{1}{2} sin (2 arcsin (sin (x))))

Solution steps

\int_{0}^{s i n (x)} 1 - t^{2} d t

Apply Trigonometric Substitution

= \int_{0}^{a r c s i n (s i n (x))} cos^{2} (u) d u

Rewrite using trig identities

= \int_{0}^{a r c s i n (s i n (x))} \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{a r c s i n (s i n (x))} 1 + cos (2 u) 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

= \frac{1}{2} (\int_{0}^{a r c s i n (s i n (x))} 1 d u + \int_{0}^{a r c s i n (s i n (x))} cos (2 u) d u)

\int_{0}^{a r c s i n (s i n (x))} 1 d u = arcsin (sin (x))

\int_{0}^{a r c s i n (s i n (x))} cos (2 u) d u = \frac{1}{2} sin (2 arcsin (sin (x)))

= \frac{1}{2} (arcsin (sin (x)) + \frac{1}{2} sin (2 arcsin (sin (x))))

\int_{0}^{π} 3 e^{x} - 2 sin (x) d x

\int_{0}^{2} x ∣ x - 1 ∣ d x

\int_{0}^{1} e^{x} 1 - x d x

\int_{0}^{2} e^{x^{2} - x} d x

\int_{0}^{y} \frac{1}{y} d y

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