How do you get the integral?

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

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

  • 2
    $\begingroup$ 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

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


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

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


$\displaystyle=4+2\sin\theta+4\cos2\theta+2\sin3\theta$ (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 $

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.