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So we have this identity : $$\forall\alpha\in\left(0,1\right),\int_{0}^{\pi/2}\frac{\sin^2(\theta)}{\left(1-\alpha^2\sin^2(\theta)\right)^{\frac{3}{2}}}d\theta=\int_{0}^{\pi/2}\frac{\cos^2(\theta)}{\left(1-\alpha^2\right)\sqrt{1-\alpha^2\sin^2(\theta)}}d\theta$$ I have no clue as of how to prove it.

The thing is, I actually need to generalize it : $$\forall p \in\mathbb{N}^*,\forall\alpha\in\left(0,1\right),\int_{0}^{\pi/2}\frac{\sin^p(\theta)}{\left(1-\alpha^p\sin^p(\theta)\right)^{\frac{3}{2}}}d\theta=\text{ }?$$ But in order to tackle the generalization, the first step is obviously to understand how to prove the case $p=2$ above, in order to adapt it and thus find a suitable expression for the generalized RHS.

So, any ideas for proving the first identity ?

Harmonic Sun
  • 2,014

2 Answers2

2

$\alpha\in ]0;1[$, perform integration by parts,

\begin{align}\int_0^{\frac{\pi}{2}}\frac{\sin^2 t}{\left(1-\alpha^2\sin^2t\right)^{\frac{3}{2}}}\,dt&=\left[-\frac{\cos t}{(1-\alpha^2)\sqrt{1-\alpha^2\sin^2 t}}\times \sin t\right]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}}\frac{\cos^2 t}{(1-\alpha^2)\sqrt{1-\alpha^2\sin^2 t}}\\ &=\int_0^{\frac{\pi}{2}}\frac{\cos^2 t}{(1-\alpha^2)\sqrt{1-\alpha^2\sin^2 t}} \end{align}

NB:

\begin{align}\frac{\partial }{\partial t}\frac{-\cos t}{\sqrt{1-\alpha^2\sin^2 t}}&=\frac{(\sin t)\sqrt{1-\alpha^2\sin^2 t}-2\alpha^2\cos^2 t\sin t\times \frac{1}{2\sqrt{1-\alpha^2\sin^2 t}}}{1-\alpha^2\sin^2 t}\\ &=\frac{(\sin t)\Big(1-\alpha^2\sin^2 t-\alpha^2\cos^2 t\Big)}{\Big(1-\alpha^2\sin^2 t\Big)\times \Big(1-\alpha^2\sin^2 t\Big)^{\frac{1}{2}}}\\ &=\frac{(\sin t)\Big(1-\alpha^2\Big)}{\Big(1-\alpha^2\sin^2 t\Big)^{\frac{3}{2}}}\\ \end{align}

FDP
  • 15,643
1

Let the integral on LHS be denoted by $I(a) $ so that $$I(a) =\int _{0}^{\pi/2}\frac{\sin^2t}{(1-a^2\sin^2t)^{3/2}}\,dt=\frac{1}{a^2}\int_{0}^{\pi/2}\frac{dt}{(1-a^2\sin^2t)^{3/2}}-\frac{1}{a^2}\int_{0}^{\pi/2}\frac{dt}{\sqrt{1-a^2\sin^2t}}\tag{1}$$ Now note that $$\frac{d} {dt} \frac{\sin t\cos t} {\sqrt{1-a^2\sin^2t} } =\frac{1}{a^2}\sqrt{1-a^2\sin^2t}-\frac{1-a^2}{a^2}\cdot\frac{1}{(1-a^2\sin^2t)^{3/2}}$$ Integrating the above on interval $[0,\pi/2]$ we get $$\int_{0}^{\pi/2}\frac{dt}{(1-a^2\sin^2t)^{3/2}}=\frac{1}{1-a^2}\int _{0}^{\pi/2}\sqrt{1-a^2\sin ^2t}\,dt\tag{2}$$ Using $(2)$ in $(1)$ we get $$I(a) =\frac{1}{a^2(1-a^2)}\int_{0}^{\pi/2}\sqrt {1-a^2\sin^2t}\,dt-\frac{1}{a^2}\int_{0}^{\pi/2}\frac{dt}{\sqrt {1-a^2\sin^2t}}=\frac{1}{1-a^2}\int_{0}^{\pi/2}\frac{\cos^2t}{\sqrt{1-a^2\sin^2t}}\,dt$$ The symbols $a, t$ have been used in place of $\alpha, \theta$ to reduce typing effort.

Paramanand Singh
  • 92,128