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I was not able to find the value of the integral

$$\int_{0}^{\pi/2}\frac{\ln(1+3\sin^2x)}{\sin^2x}dx$$

So, I refered to the solution provided as follows. $$f(t)=\int_{0}^{\pi/2}\frac{\ln(1+t\sin^2x)}{\sin^2x}dx$$ $$\begin{align*} f'(t)&=\int_{0}^{\pi/2}\frac{dx}{1+t\sin^2x}\\ &=\int_{0}^{\pi/2}\frac{\text{cosec}^2x}{\cot^2x+1+t}dx\\ &=\int_{0}^{\infty}\frac{dp}{p^2+1+t}\\ &=\left[\frac 1{\sqrt{1+t}}\tan^{-1}{\frac p{\sqrt{1+t}}}\right]_0^{\infty}=\frac{\pi}{2{\sqrt{1+t}}}\end{align*} $$

$$f(t)=\int f'(t)dt=\int \frac{\pi}{2{\sqrt{1+t}}}dt =\pi{\sqrt{1+t}}+C$$ $$f(0)=0 \implies C=-\pi$$ $$\implies f(t)=\pi(\sqrt{1+t}-1)$$ Therefore, the original integral is $$\int_{0}^{\pi/2}\frac{\ln(1+3\sin^2x)}{\sin^2x}dx=f(3)=(\sqrt 4-1)\pi=\pi$$

This method is not obvious to me. Can someone provide an alternative solution?

Thanks in advance.

Quanto
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Asher2211
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    This method is called differentiating under the integral sign(Feynman's technique of integration).It is very useful and powerful method and is generally not taught in calculus 1 classes but it is far more common in higher courses.

    Following integral can also be integrated using above method(try it!) $$I=\int_{0}^{1} \frac{\sin(ln\space x)}{ln(x)}dx$$

     – Paras Jul 14 '21 at 19:01
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    @Paras Thanks for introducing me to Feynman's technique. Are there any other similar techniques which are not popular. – Asher2211 Jul 14 '21 at 19:07
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    No, This technique is popular but not among the peopls who are taking basic calculus course. I think you should look up integrals on this site which suits your level because they will surely teach you techniques that are not generally found in textbooks. – Paras Jul 14 '21 at 19:12
  • @Paras Can you name some of them – Asher2211 Jul 14 '21 at 19:13
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    As far as basic integration is concerned I can't recall any such technique with a specific name as in feynman's technique case. You should look up Question on this site which will teach you the best method to calculate an integral, not the short-cuts(they are sometimes temporary) – Paras Jul 14 '21 at 19:20
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    If you want more applications of feynman technique, you can look up answers of @Quanto because many of his answers uses this technique. – Paras Jul 14 '21 at 19:41
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    https://math.stackexchange.com/questions/3453227/when-to-differentiate-under-the-integral-sign/3453393#3453393 – Quanto Jul 15 '21 at 14:13
  • @Quanto Thank you for the link. I was trying to use Feynman technique to solve the integral $\displaystyle\int_0^{\pi/2}{\ln(\sin x)}dx$. I defined $\displaystyle f(t)=\int_0^{\pi/2}{\ln(t\sin x)}dx$ and after solving got $\displaystyle f(t)=\frac{\pi}{2}\ln t+C$, but I am not able to find the consatant . Can we use Feymans technique here? – Asher2211 Jul 16 '21 at 05:46
  • @Asher2211 - The Faynmann trick isn’t suitable for just any integral. It’d be awkward for this one. – Quanto Jul 16 '21 at 21:17

2 Answers2

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Alternatively

\begin{align} \int_{0}^{\pi/2}\frac{\ln(1+3\sin^2x)}{\sin^2x}dx= &\int_{0}^{\pi/2}\frac{\ln(1+3\cos^2x)}{\cos^2x}dx = \int_{0}^{\pi/2}{\ln(1+3\cos^2x)}\>d(\tan x)\\ \overset{IBP} =&\int_{0}^{\pi/2}\frac{6\sin^2x}{1+3\cos^2x}dx = \int_{0}^{\pi/2}\left( \frac{8}{1+3\cos^2x} -2\right)dx\\ =&8\int_{0}^{\pi/2} \frac{d(\tan x)}{4+\tan^2x}-\pi = 2\pi -\pi=\pi \end{align}

Quanto
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Using the series of $\log(1+x)$ for all $x\in[0,1]$, the integral $$\int_0^{\frac{\pi}{2}}\frac{\log(1+\sin^2 x)}{\sin ^2 x} dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\int_0^{\frac{\pi}{2}}3^{n+1}\sin^{2n}xdx=\frac{\pi}{2}\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\left(\frac{3}{4}\right)^{n+1}\left[{2n\choose n}\right]$$ since the Wallis' Integral $\int_0^{\frac{\pi}{2}}\sin^{2n}xdx=\frac{\pi}{24^n}{2n\choose n}$. To evaluate the latter expression we use the ordinary generating function of central coefficients, namely $\sum_{n=0}^{\infty}{2n\choose n}\left(-\frac{x}{4}\right)^n=\frac{1}{\sqrt{1+x}}$ for $|x|<1$ On integrating from $x=0$ to $x=3$ we get $$\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\left(\frac{3}{4}\right)^{n+1}{2n\choose n}=\int_0^3\frac{dx}{\sqrt {1+x}}=4-2=2$$. On multiplying both sides by $\frac{\pi}{2}$, the required result of integral is $\pi$.

Naren
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