What is the integral from 0 to sin(2x) of arcsin(t) ?

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

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integral from 0 to sin(2x) of arcsin(t)

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

cos (2 x) + sin (2 x) arcsin (sin (2 x)) - 1

Solution steps

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

Apply Integration By Parts

: [t arcsin (t) - \int \frac{t}{1 - t ^{2}} d t]_{0}^{s i n (2 x)}

= [t arcsin (t) - \int \frac{t}{1 - t ^{2}} d t]_{0}^{s i n (2 x)}

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

= [t arcsin (t) - (- 1 - t^{2})]_{0}^{s i n (2 x)}

Simplify

= [t arcsin (t) + 1 - t^{2}]_{0}^{s i n (2 x)}

Compute the boundaries:

cos (2 x) + sin (2 x) arcsin (sin (2 x)) - 1

= cos (2 x) + sin (2 x) arcsin (sin (2 x)) - 1

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

\int_{x}^{\infty} e^{- x} d x

\int_{1}^{3} x^{2} - 6 x + 5 d x

\int_{0}^{4} \frac{1}{4 x ^{2} + 9} d x

\int_{2}^{3} x^{2} + x - 2 d x

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