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How do you get the integral?

$$\int x \sqrt{4x-x^2} dx$$

I have no idea on how to integrate it. Thank you.

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    $\begingroup$ Complete the square and then try to see if you can apply integration methods such as trig substitution. A $u$-substitution might help. $\endgroup$ – ZanCoul Nov 23 '13 at 6:41
  • $\begingroup$ Definitely try completing the square and use this hint: $tan^2(\theta) + 1 = sec^2(\theta)$, when thinking about substitution. $\endgroup$ – InsigMath Nov 23 '13 at 6:47
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As $\displaystyle 4x-x^2=4-(x-2)^2$ following this set $x-2=2\sin\theta\ \ \ \ (1)$

Due to the reason explained here, $\displaystyle\sqrt{4x-x^2}=\sqrt{4\cos^2\theta}=+2\cos\theta\ \ \ \ (2)$

So, $\displaystyle\int x\sqrt{4x-x^2}dx=\int(2\sin\theta+2)2\cos\theta\cdot 2\cos\theta d\theta$

$\displaystyle=4\int(1+\sin\theta)(2\cos^2\theta)d\theta$

Now, $\displaystyle4(1+\sin\theta)(2\cos^2\theta)$

$\displaystyle=4(1+\sin\theta)(1+\cos2\theta)$ (using $\cos2A=2\cos^2A-1$)

$\displaystyle=4+4\sin\theta+4\cos2\theta+4\sin\theta\cos2\theta$

$\displaystyle=4+2\sin\theta+4\cos2\theta+2\sin3\theta$ (using $2\sin B\cos A=\sin(A+B)-\sin(A-B)$)

We need to use $\displaystyle\cos mxdx=\frac{\sin mx}m+C$ and $\displaystyle\sin mxdx=-\frac{\cos mx}m+K$

From $(1),(2);$ we already have $\displaystyle\sin\theta=\frac{x-2}2\implies\theta=\arcsin\frac{(x-2)}2$ and $\displaystyle\cos\theta=+\frac{\sqrt{4x-x^2}}2 $

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  • $\begingroup$ According to my little knoldge Ans of lab bhattacharjee has an small error for solving Aove function we have to take x commom then x=4sin^t than we get root of 4x-x^2=2sint then by simple manipulation corresponding integration will solved $\endgroup$ – Gaurav.N.Pal Nov 23 '13 at 16:06
  • $\begingroup$ @Gaurav.N.Pal. I've set $x-2=2\sin\theta$ $$\implies 4x-x^2=4-(x-2)^2=4-(2\sin\theta)^2=4\cos^2\theta$$. Can you please pinpoint the mistake? $\endgroup$ – lab bhattacharjee Nov 24 '13 at 2:48

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