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$$\int_{0}^{a} \frac{x \cdot dx}{x+\sqrt{a^2-x^2}}$$ This is what I tried:

Let $x=a\cdot \sin(t)$

$$\int_{0}^{a} \frac{x\cdot dx}{x+\sqrt{a^2-x^2}} = \int_{0}^{\pi/2} \frac{a\sin(t)\cdot a\cos(t)dt}{a\sin(t)+a\cos(t)} = a\cdot\int_{0}^{\pi/2}\frac{\sin(t)\cdot \cos(t)dt}{\sin(t)+\cos(t)}$$

I can't this of a way to integrate this

So Lo
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    you can try $\sin t \cos t=\frac{(\sin t +\cos t)^2-1}{2}$. – Riemann Jun 10 '18 at 08:26
  • There are many ways to solve this integral. Personally, I would not go with trigonometric substitution. Rationalize the fraction to get rid of the root in the denominator. Then, use linearity. If you do want to stick with a trigonometric substitution, you might want to give this a look. – an4s Jun 10 '18 at 08:28

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Hint. We have that $$\int_{0}^{\pi/2}\frac{\sin(t)\cdot \cos(t)}{\sin(t)+\cos(t)}dt= \frac{1}{2}\int_{0}^{\pi/2}\frac{(\sin(t)+\cos(t))^2}{\sin(t)+\cos(t)} -\frac{1}{2}\int_{0}^{\pi/2}\frac{dt}{\sin(t)+\cos(t)}.$$ For the last integral, note that $\sin(t)+\cos(t)=\sqrt{2}\cos(t-\pi/4)$ and see How to integrate $\int \frac{1}{\cos(x)}\,\mathrm dx$

Robert Z
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