What is the integral from 1/4 to 1 of arcsin(x) ?

The integral from 1/4 to 1 of arcsin(x) is pi/2-(arcsin(1/4)+sqrt(15))/4

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integral from 1/4 to 1 of arcsin(x)

\int_{\frac{1}{4}}^{1} arcsin (x) d x

\frac{π}{2} - \frac{arcsin ( \frac{1}{4} ) + 15}{4}

Decimal Notation

0.53938 \dots

Solution steps

\int_{\frac{1}{4}}^{1} arcsin (x) d x

Apply Integration By Parts

: [x arcsin (x) - \int \frac{x}{1 - x ^{2}} d x]_{\frac{1}{4}}^{1}

= [x arcsin (x) - \int \frac{x}{1 - x ^{2}} d x]_{\frac{1}{4}}^{1}

\int \frac{x}{1 - x ^{2}} d x = - 1 - x^{2}

= [x arcsin (x) - (- 1 - x^{2})]_{\frac{1}{4}}^{1}

Simplify

= [x arcsin (x) + 1 - x^{2}]_{\frac{1}{4}}^{1}

Compute the boundaries:

\frac{π}{2} - \frac{arcsin ( \frac{1}{4} ) + 15}{4}

= \frac{π}{2} - \frac{arcsin ( \frac{1}{4} ) + 15}{4}

\int_{π}^{0} (sin (x) - 2)^{2} d x

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

\int_{0}^{16} \frac{x}{4 + 2 x} d x

\int_{7}^{14} 0.05 e^{- 0.05 x} d x

\int_{- \infty}^{0} e^{\frac{x}{4}} d x

What is the integral from 1/4 to 1 of arcsin(x) ?
The integral from 1/4 to 1 of arcsin(x) is pi/2-(arcsin(1/4)+sqrt(15))/4