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Please answer this question​

Please answer this question​-example-1
User Brydgesk
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For the integral ∫asin(x)dx, use integration by parts ∫udv=uv−∫vdu.

Let u=asin(x) and dv=dx.

Then du=(asin(x))′dx=
\rm \frac{dx}{ \sqrt{1 - {x}^(2) } } and v=∫1dx=x

So,


\rm{\int{\operatorname{asin}{\left(x \right)} d x}}=\color{h}{\left(\operatorname{asin}{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{\sqrt{1 - x^(2)}} d x}\right)}=\color{h}{\left(x \operatorname{asin}{\left(x \right)} - \int{\frac{x}{\sqrt{1 - x^(2)}} d x}\right)} \\

Let u=1−x2.

Then du=(1−x2)′dx=−2xdx and we have that xdx=−du/2.

The integral can be rewritten as


\rm x \operatorname{asin}{\left(x \right)} - \color{g}{\int{\frac{x}{\sqrt{1 - x^(2)}} d x}} = x \operatorname{asin}{\left(x \right)} - \color{j}{\int{\left(- (1)/(2 √(u))\right)d u}} \\

Apply the constant multiple rule ∫cf(u)du=c∫f(u)du with c=-1/2 and f(u)=
(1)/( √(u) )


\rm x \operatorname{asin}{\left(x \right)} - \color{h}{\int{\left(- (1)/(2 √(u))\right)d u}} = x \operatorname{asin}{\left(x \right)} - \color{h}{\left(- \frac{\int{(1)/(√(u)) d u}}{2}\right)} \\

Apply the power rule


\rm\int u^(n)\, du = (u^(n + 1))/(n + 1) \\

with n=−1/2


\rm x \operatorname{asin}{\left(x \right)} + \frac{\color{h}{\int{(1)/(√(u)) d u}}}{2}=x \operatorname{asin}{\left(x \right)} + \frac{\color{j}{\int{u^{- (1)/(2)} d u}}}{2}=x \operatorname{asin}{\left(x \right)} + \frac{\color{j}{\frac{u^{- (1)/(2) + 1}}{- (1)/(2) + 1}}}{2}=x \operatorname{asin}{\left(x \right)} + \frac{\color{j}{\left(2 u^{(1)/(2)}\right)}}{2}=x \operatorname{asin}{\left(x \right)} + \frac{\color{h}{\left(2 √(u)\right)}}{2} \\


\rm Recall \: that  \: u=1− {x}^(2)


\rm x \operatorname{asin}{\left(x \right)} + \sqrt{\color{re}{u}} = x \operatorname{asin}{\left(x \right)} + \sqrt{\color{rd}{\left(1 - x^(2)\right)}} \\

Therefore,


\rm\int{\operatorname{asin}{\left(x \right)} d x} = x \operatorname{asin}{\left(x \right)} + \sqrt{1 - x^(2)}+C \\

User Jussi Kukkonen
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