Solve the indefinite integral

Question

I found a paper online which had integral problems.

This was one of them and it was the only one which I couldn't solve. I tried to use by-parts, u-subsititution, but failed.

So please help me out of this. This may seem easy to some but please I am just a beginner. Here is the question:

$$\int \sqrt{2x - x^2}dx$$

Thanks in advance.

Answer 1

The hint:

$2x-x^2=1-(x-1)^2$ and substitute $x-1=\cos{t}$, where $0\leq t\leq\pi$;

use $$\sqrt{1-\cos^2t}=|\sin{t}|=\sin{t}$$ and $$\sin^2t=\frac{1-\cos2t}{2}.$$ Can you end it now?

Answer 2

$$2x - x^2 = - (x^2-2x) = - (x^2-2x+1) +1 = 1-(x-1)^2$$

Now let $u=x-1$

And use the formula $$\int\sqrt{1-u^2}du = \frac u2\sqrt{1-u^2} + \frac 1 2\arcsin u+c$$

Answer 3

Note that $$\int \sqrt{2x-x^2}dx=\int \sqrt{1-(x-1)^2}dx$$Thus, it is convenient to substitute $x-1=\cos{u}$ to get $\frac{dx}{du}=-\sin u\implies dx=-\sin (u)du $. Hence, $$\int \sqrt{1-(x-1)^2}dx=-\int\sqrt{1-\cos^2u}\cdot\sin(u)~du=\int \sin^2u~du$$Now use the identity that $\sin^2u=\frac{-cos(2u)+1}{2}$ to get, $$ -\int\frac{-cos(2u)+1}{2}du=\frac{1}{2}\int\cos(2u)du-\frac{1}2\int du={\frac{1}2\left[\frac{sin(2u)}{2}-u\right]+C} $$ Now, since $u=\arcsin(x-1)$, we get the result equal to $$\boxed{\frac{1}2\left[\frac{\sin(2\arcsin(x-1))}{2}-u\right]+C}$$

Answer 4

Hint: Use the so called Euler substitution: $$\sqrt{(-1)(x-0)(x-2)}=(x-2)\cdot t$$