# Transformation to $\log\left(\frac a2\cos x\right)$

Question: How does$$T=-\int\limits_{\tfrac {\pi}2}^{-\tfrac {\pi}2}dx\,\frac {\log\left(\tfrac {a^2}4\cos^2x\right)}{\sqrt{\tfrac {a^2}4\cos^2x}}\left(\frac a2\cos x\right)=4\int\limits_0^{\tfrac {\pi}2}dx\,\log\left(\frac a2\cos x\right)\tag1$$

I was evaluating an integral, namely$$I=\int\limits_0^adx\,\frac {\log x}{\sqrt{ax-x^2}}$$And I'm having trouble seeing how to get from the left-hand side to the right-hand side. Doesn't the denominator and numerator cancel to leave you with$$T=-\int\limits_{\tfrac {\pi}2}^{-\tfrac {\pi}2}dx\,\log\left(\frac {a^2}4\cos^2x\right)=-2\int\limits_{\tfrac {\pi}2}^{-\tfrac {\pi}2}dx\,\log\left(\frac a2\cos x\right)$$But I don't see how that can be substituted to get the second expression of $(1)$. What kind of substitution should be made to make the transformation from where I left off to $(1)$?

• The function is an even function and for even functions $\displaystyle \int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx$ – MathNovice Nov 21 '17 at 4:06
• @MathNovice Oh crap! I completely forgot that rule! What's the rule for an odd function then? – Crescendo Nov 21 '17 at 4:10
• If $f(x)$ is an odd function, same integral goes to zero. – MathNovice Nov 21 '17 at 4:13
• Don't you know that $\sqrt{t^2}=\mid t\mid$? – FDP Nov 22 '17 at 10:43

First, $\cos(x)\ge 0$ and is even for $x\in[-\pi/2,\pi/2]$. Hence, $\sqrt{\cos^2(x)}=\cos(x)$. If $a>0$, then $\sqrt{a^2}=a$. Therefore,
$$\frac{\log\left(\frac{a^2}{4}\cos^2(x)\right)\left(\frac a2\cos(x)\right)}{\sqrt{\frac{a^2}{4}\cos^2(x)}}=\frac{\log\left(\left(\frac{a\cos(x)}{2}\right)^2\right)\left(\frac a2\cos(x)\right)}{\frac a2\cos(x)}=2\log\left(\frac{a\cos(x)}{2}\right)$$
\begin{align} \int_{-\pi/2}^{\pi/2}\frac{\log\left(\frac{a^2}{4}\cos^2(x)\right)\left(\frac a2\cos(x)\right)}{\sqrt{\frac{a^2}{4}\cos^2(x)}}\,dx&=\int_{-\pi/2}^{\pi/2}2\log\left(\frac{a\cos(x)}{2}\right)\,dx\\\\ &=4\int_0^{\pi/2}\log\left(\frac{a\cos(x)}{2}\right)\,dx \end{align}
• If you don't mind, can you also tell me how you might see that the substitution $x=\tfrac a2(1-\sin t)$ be the right substitution from the original integral$$\int\limits_0^{a}dx\,\frac {\log(ax-x^2)}{\sqrt{ax-x^2}}$$? I'm having trouble seeing how someone might get there. – Crescendo Mar 31 '18 at 4:14