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

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

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

\int_{c o s (x)}^{1} 1 - t^{2} d t

\frac{1}{2} (\frac{π}{2} - arcsin (cos (x)) - \frac{1}{2} sin (2 arcsin (cos (x))))

Solution steps

\int_{c o s (x)}^{1} 1 - t^{2} d t

Apply Trigonometric Substitution

: \int_{a r c s i n (c o s (x))}^{\frac{π}{2}} cos^{2} (u) d u

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

Rewrite using trig identities

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

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

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

= \frac{1}{2} (\frac{π}{2} - arcsin (cos (x)) - \frac{1}{2} sin (2 arcsin (cos (x))))

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