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

The integral from 0 to sqrt(1-z^2 of)zsin(x) is zsin(x)sqrt(-z^2+1)

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integral from 0 to sqrt(1-z^2 of)zsin(x)

\int_{0}^{1 - z^{2}} z sin (x) d y

z sin (x) - z^{2} + 1

Solution steps

\int_{0}^{1 - z^{2}} z sin (x) d y

Integral of a constant:

\int a d x = a x

= [sin (x) zy]_{0}^{1 - z^{2}}

Compute the boundaries:

z sin (x) - z^{2} + 1

= z sin (x) - z^{2} + 1

\int_{- 1}^{2.5} f (t) d t

\int_{0}^{2} 8 e^{2 t} d t

\int_{- 3}^{5} x^{2} - 2 x - 15 d x

\int_{1}^{3} (x^{2} - 4 x) d x

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

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