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Calculate the following integral $$I(a)=\int_0^{\pi/2}\ln(a^2-\sin^2x)dx$$

My attempt: $$I'(a)=\int_0^{\pi/2}\dfrac{2a}{a^2-\sin^2x}dx=\int_0^{\pi/2}\left(\dfrac{1}{a+\sin x}+\dfrac{1}{a-\sin x}\right)dx$$ I don't know how to continue though :( Could someone help me or have any idea?

Sine of the Time
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Alex Nguyen
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  • Let's continue your work. We can apply Weiestrass substitution on the first integral (so is the second one), that is, let $t=\tan\dfrac{x}{2}$. Then $$dt=\dfrac{1}{2}\sec^2\dfrac{x}{2}dx=\dfrac{1+t^2}{2}dx$$ Thus the integral becomes $$\int^{\pi/2}_0\dfrac{1}{a+\dfrac{2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2\frac{x}{2}+\sin^2\frac{x}{2}}}dx=\int^1_0\dfrac{1}{a+\dfrac{2t}{1+t^2}}\dfrac{2}{1+t^2}dt=\int^1_0\dfrac{2}{at^2+2t+a}dt$$ This is a quadratic integral. You need to classify the cases into $2^2-4a^2<0,=0$ and $>0$ respectively. Even though summing up two integrals didnt give good way. – Angae MT Aug 21 '24 at 15:25
  • which values can $a$ take? – Sine of the Time Aug 21 '24 at 15:33
  • @SineoftheTime $a>1$ of course! – Harry Aug 21 '24 at 15:49
  • Related – Sine of the Time Aug 21 '24 at 15:49
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    $\int_0^{\frac{\pi}{2}} \frac{1}{a+b\sin x},dx=\frac{1}{\sqrt{b^2-a^2}}\left[{\ln\left(\frac{\left|\sqrt{b^{2} - a^{2}} - b - a\right|}{\left|\sqrt{b^{2} - a^{2}} + b + a\right|}\right)} - {\ln\left(\frac{\left|2 \sqrt{b^{2} - a^{2}} - 2b\right|}{\left|2 \sqrt{b^{2} - a^{2}} + 2b\right|}\right)}\right]$ – Amrut Ayan Aug 21 '24 at 16:25
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    $$I(a)=\int_0^{\pi/2}\left(\ln(\cos^2x)+\ln(\frac{a^2}{\cos^2x}-\tan^2x)\right)dx$$ $$\overset{t=\tan x}{=}-\pi\ln2+\int_0^\infty\frac{\ln\big(a^2+(a^2-1)t^2\big)}{1+t^2}dt=\pi\ln(a+\sqrt{a^2-1})-\pi\ln2$$ – Svyatoslav Aug 21 '24 at 16:27
  • @Svyatoslav How you compute $\int_0^{\infty}\dfrac{\ln (a^2+(a^2-1)t^2)}{1+t^2}dt$? – Harry Aug 22 '24 at 00:55
  • $$\int_0^\infty\frac{\ln\big(a^2+(a^2-1)t^2\big)}{1+t^2}dt=\Re\int_{-\infty}^\infty\frac{\ln\big(a-i\sqrt{a^2-1},t\big)}{(t+i)(t-i)}dt$$ Closing the contour in the upper half-plane $$=\Re,2\pi i,\underset{z=i}{Res}\frac{\ln\big(a-i\sqrt{a^2-1},z\big)}{(z+i)(z-i)}=\Re,2\pi i\frac{\ln\big(a+\sqrt{a^2-1}\big)}{2i}=\pi\ln\big(a+\sqrt{a^2-1}\big)$$ – Svyatoslav Aug 22 '24 at 02:39

4 Answers4

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Here’s my continuation to your last step, $$\int_0^{\frac{\pi}{2}} \frac{1}{a+b\sin x}\,dx=\frac{1}{\sqrt{b^2-a^2}}\left[{\ln\left(\frac{\left|\sqrt{b^{2} - a^{2}} - b - a\right|}{\left|\sqrt{b^{2} - a^{2}} + b + a\right|}\right)} - {\ln\left(\frac{\left|2 \sqrt{b^{2} - a^{2}} - 2b\right|}{\left|2 \sqrt{b^{2} - a^{2}} + 2b\right|}\right)}\right]$$

In your first integral, set $b=1$

$$I_1=\int_0^{\frac{\pi}{2}} \frac{1}{a+\sin x}\,dx=\frac{1}{\sqrt{1-a^2}}\left[{\ln\left(\frac{\left|\sqrt{1 - a^{2}} - (1+a)\right|}{\left|\sqrt{1 - a^{2}} + (1 + a)\right|}\right)} - {\ln\left(\frac{\left| \sqrt{1 - a^{2}} - 1\right|}{\left| \sqrt{1 - a^{2}} + 1\right|}\right)}\right]$$

In your second integral, set $b=-1$

$$I_2=\int_0^{\frac{\pi}{2}} \frac{1}{a-\sin x}\,dx=\frac{1}{\sqrt{1-a^2}}\left[{\ln\left(\frac{\left|\sqrt{1 - a^{2}} + 1 - a\right|}{\left|\sqrt{1 - a^{2}}+a-1\right|}\right)} - {\ln\left(\frac{\left| \sqrt{1- a^{2}} + 1\right|}{\left| \sqrt{1 - a^{2}} - 1\right|}\right)}\right]$$

$$\frac{\partial I(a)}{\partial a}=I_1+I_2$$

(I have not given proof to above generalized integral as it’s pretty popular and doable using the half angle tangent substitution and some partial fractions)

But i think this is too messy to solve, if you are interested to carry on, refer here

Anyways, I found a nice closed form to share, i will leave it for reference; $$\boxed{\int_0^{\pi/2}\frac{1}{1+a\sin x }\, dx=\frac{2}{\sqrt{1-a^2}}\left(\tan^{-1}\left(\sqrt{\frac{1+a}{1-a}}\right) - \tan^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)\right)}$$

Here's the easier alternative;

Consider the integral from here

Set $a^2=A$ and $b^2=A-1$

We get, $$J(a)=\int_0^{\pi/2}\log(A-\sin^2(x))dx=\pi\ln \left(\frac{\sqrt{A}+\sqrt{A-1}}{2}\right)$$

So for $a>1$, $$\boxed{I(a)=\int_0^{\pi/2}\ln(a^2-\sin^2x)\,\mathrm dx=\pi\ln \left(\frac{{a}+\sqrt{a^2-1}}{2}\right)}$$

Amrut Ayan
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    $$(1)\quad\tan^{-1}\sqrt{\frac{1+a}{1-a}}=\frac\pi4+\frac12\arcsin a\(2)\quad\tan^{-1}\frac a{\sqrt{1-a^2}}=\arcsin a\\implies(1)-(2)=\frac\pi4-\frac12\arcsin a=\frac12\arccos a$$ – user170231 Aug 23 '24 at 18:01
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Note $$\begin{eqnarray} a^2-\sin^2x&=&a^2-\frac12(1-\cos(2x))=a^2-\frac12+\frac12\cos(2x)\\ &=&|A+B(\cos(2x)+i\sin(2x))|^2\\ &=&|A+Be^{2ix}|^2 \end{eqnarray}$$ and hence $$\begin{eqnarray} \ln(a^2-\sin^2x)&=&2\ln|A+Be^{2ix}|=2\Re\ln(A+Be^{2ix}). \end{eqnarray}$$ Here $$ A= \frac{1}{2} \left(\sqrt{a^2-1}+a\right), B= \frac{1}{2} \left(\sqrt{a^2-1}-a\right). $$ So $$\begin{eqnarray} I(a)&=&\int_0^{\pi/2}\ln(a^2-\sin^2x)\,\mathrm dx\\ &=&\frac12\int_{-\pi/2}^{\pi/2}\ln(a^2-\sin^2x)\,\mathrm dx\\ &=&\Re\int_{-\pi/2}^{\pi/2}\ln(A+Be^{2ix})\,\mathrm dx\\ &\overset{z=e^{2ix}}=&\frac12\Re\int_{|z|=1}\ln(A+Bz)\frac{dz}{iz}\\ &=&\frac1{2}\Re\frac1i\int_{|z|=1}\frac{\ln(A+Bz)}{z}dz\\ &=&\frac1{2}\cdot2\pi\ln A=\pi\ln A. \end{eqnarray}$$

xpaul
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Differentiating w.r.t. $a$ (where $a\ge 1$)

$$ I(a)=\int_0^{\frac{\pi}{2}} \ln \left(a^2-\sin ^2 x\right) d x $$ yields $$ \begin{aligned} I^{\prime}(a) & =2 a \int_0^{\frac{\pi}{2}} \frac{1}{a^2-\sin ^2 x} d x \\ & =2 a \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{a^2 \sec ^2 x-\tan ^2 x} d x \\ & =2 a \int_0^{\infty} \frac{d t}{a^2+\left(a^2-1\right) t^2}, \quad t=\tan x \\ & =\frac{2}{\sqrt{a^2-1}}\left[\tan ^{-1}\left(\frac{t\sqrt{a^2-1} }{a}\right)\right]_0^{\infty} \\ & =\frac{\pi}{\sqrt{a^2-1}} \end{aligned} $$ Integrating back from $x=1$ to $a$ yields $$ \begin{aligned} &I(a)-I(1)=\int_1^a \frac{\pi}{\sqrt{x^2-1}} d x \\ \end{aligned} $$ Rearranging gives

$$ \begin{aligned} I(a)&=\frac{\pi}{2}\left[\ln \left(\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}-x}\right)\right]_1^a+\int_0^{\frac{\pi}{2}} \ln \left(\cos^2 x\right) dx \\& =\frac{\pi}{2} \ln \left(\frac{\sqrt{a^2-1}+a}{\sqrt{a^2-1}-a}\right)+\pi \ln 2 \\ &=\pi \ln \left(\frac{\sqrt{a^2-1}+a}{2}\right) \end{aligned} $$

Lai
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Note that $\int_0^{\pi/2}\frac{dx}{a\pm \sin x}=\frac{\sec^{-1}(\pm a)}{\sqrt{a^2-1}}$

\begin{align} I'(a)&=\int_0^{\pi/2}\left(\dfrac{1}{a+\sin x}+\dfrac{1}{a-\sin x}\right)dx\\ &= \frac{\sec^{-1}(a)+ \sec^{-1}(-a)}{\sqrt{a^2-1}}=\frac\pi{\sqrt{a^2-1}}=(\pi\cosh^{-1}a)’\\ I(a)&= I(1) +\int_1^a I’(t)dt= - \pi\ln2+\pi\cosh^{-1}a \end{align}

Quanto
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