Willing to evaluate $$\int_{0}^{l-a} \frac{dx}{\sqrt{(a+x)^2-a^2} }$$
The answer is given $ \cosh ^{-1} (l/a)$. Don't know how this can be achieved.
Please help
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Note that once you have the answer, you can check it by taking the derivative of both expressions with respect to $l$ (and checking the equality at one point—probably $l=a$ is best). – Greg Martin May 17 '18 at 07:00
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Hint: Substitute $$\sqrt{x^2+2ax}=x+t$$ then we get $$x=\frac{t^2}{2a-2t}$$ and $$dx=\frac{4at-2t^2}{(2a-2t)^2}dt$$ with the second line you Can express the term with the square root as a function in $t$ $$\sqrt{x^2+2ax}=t+\frac{t^2}{2a-2t}$$ $$\sqrt{x^2+2ax}=1/2\,{\frac {t \left( 2\,a-t \right) }{a-t}} $$
Dr. Sonnhard Graubner
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I am little confused about why should we assume a function of x again as a function of x and t ? – user533926 May 17 '18 at 07:01
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And what about the limit? I am getting a quadratic in t for $x=l-a$ – user533926 May 17 '18 at 07:06
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For simplification you Can solve at first the indefinite integral – Dr. Sonnhard Graubner May 17 '18 at 07:08