TL Maths

Question of the day 7.09

Tomek

·

AA HL · Integration

Find the indefinite integral:

    \[ \int \sqrt{2x-x^2}\,dx \]

Hint

First complete the square:

    \[ 2x-x^2=1-(x-1)^2. \]

Then use the substitution

    \[ x-1=\sin\theta. \]


Solution

First complete the square:

    \[ 2x-x^2=1-(x-1)^2. \]

Therefore

    \[ \int \sqrt{2x-x^2}\,dx = \int \sqrt{1-(x-1)^2}\,dx. \]

Let

    \[ x-1=\sin\theta. \]

Then

    \[ dx=\cos\theta\,d\theta. \]

Also,

    \[ \sqrt{1-(x-1)^2} = \sqrt{1-\sin^2\theta} = \cos\theta. \]

Hence

    \[ \int \sqrt{1-(x-1)^2}\,dx = \int \cos^2\theta\,d\theta. \]

Using the identity

    \[ \cos^2\theta=\frac{1+\cos 2\theta}{2}, \]

we obtain

    \[ \int \cos^2\theta\,d\theta = \frac{\theta}{2} + \frac{\sin 2\theta}{4} +C. \]

Since

    \[ \theta=\sin^{-1}(x-1), \]

and

    \[ \frac{\sin 2\theta}{4} = \frac{1}{2}\sin\theta\cos\theta = \frac{1}{2}(x-1)\sqrt{2x-x^2}, \]

the final answer is

    \[ \boxed{ \int \sqrt{2x-x^2}\,dx = \frac12(x-1)\sqrt{2x-x^2} + \frac12\sin^{-1}(x-1) +C } \]

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