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In Edwards classic treatise, p200, the following result is asked to be proved. Prove that

$$ \int_0^{a} \frac{a}{(x+\sqrt{a^2-x^2})^2}dx = \frac{1}{\sqrt{2}}\ln(1+\sqrt{2})\ $$

I have made the obvious substitution $x = a \sin\theta$ but fail to obtain the required result from integrating the resulting transformed integral.

That is, I now require to show that $$ \int_0^{\pi/2} \frac{\cos\theta}{(\cos\theta+\sin\theta)^2}\,d\theta = \frac{1}{\sqrt{2}}\ln(1+\sqrt{2})\ $$ Can anyone offer some assistance please ? I have used the tan half-angle formula but the result is not working out.

Saša
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Callie12
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1 Answers1

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Note that $\int_0^{\pi/2} \frac{\cos \theta}{(\cos\theta+\sin \theta)^2}\,d\theta =\int_0^{\pi/2} \frac{\sin \theta }{(\cos \theta+\sin \theta)^2}\,d\theta$, hence $$\int_0^{\pi/2} \frac{\cos \theta }{(\cos \theta+\sin \theta)^2}\,d\theta =\frac12\int_0^{\pi/2} \frac{1}{\cos \theta +\sin \theta }\,d\theta $$

The latter is easier to handle, for example using $\cos \theta +\sin \theta =\sqrt 2 \sin(\theta + \frac{\pi}4)$ ...

Math-fun
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