$\int \ x\sqrt{1-x^2}\,dx$, by the substitution $x= \cos t$

Question

I have tried to determine $\int \ x\sqrt{1-x^2}\,dx$ using trigonometric

formula by the substition $x=\cos t$ I have got :

$$-\int \cos t \sin^2 t\,dt\tag{1}$$ for $\sin t > 0$ and $$ \int \cos t \sin^2 t\,dt\tag{2}$$ for $\sin t <0 $.

But both $(1)$ and $(2)$ have no standard mathematical function , then how do i can determine the titled integral using this method ?

Answer 1

You can assume $t\in[0,\pi]$.

Thus, $$\sqrt{1-x^2}=|\sin{x}|=\sin{x}.$$

Answer 2

Why don’t you use the substitution $t=1-x^2$?

Then $dt=-2xdx$ and your integral becomes: $$\int x\sqrt{1-x^2}dx=-\frac 12\int \sqrt tdt=\\=-\frac{\sqrt{t^3}}3+c=\color{red}{-\frac{\sqrt{(1-x^2)^3}}3+c}$$ It’s easier in this way.

Answer 3

From $\int \cos t \sin^2 t dt$ you can use $u=\sin t, du=\cos t \ dt$ to get $\int u^2 du=\frac {u^3}3+C$ and backsubstitute.