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Evaluation of $$\int\frac{1}{(\sin x+a\sec x)^2}dx$$

Try: Let $$I=\int\frac{1}{(\sin x+a\sec x)^2}dx=\int\frac{\sec^2 x}{(\tan x+a\sec^2 x)^2}dx$$

put $\tan x=t$ and $dx=\sec^2 tdt$

So $$I=\int\frac{1}{(a+at^2+t)^2}dt$$

Could some help me how to solve above Integral , thanks in advance.

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HINT

Let $$b=\dfrac1{2a},\quad y=t+b,\tag1$$ then $$I=4b^2\int\dfrac{dt}{(t^2+2bt+1)^2} = 4b^2\int\dfrac{dy}{(y^2+1-b^2)^2} = \dfrac{4b^2}{1-b^2}\int\dfrac{(y^2+1-b^2)-y^2}{(y^2+1-b^2)^2}\,dy,$$ $$I=\dfrac{4b^2}{1-b^2}(I_1+I_2),\tag2$$ where $$I_1=\int\dfrac{dy}{y^2+1-b^2} = const + \begin{cases}\dfrac1{\sqrt{1-b^2}}\arctan\dfrac{y}{\sqrt{1-b^2}},\text{ if } b<1\\[4pt] -\dfrac1y,\text{ if }b=1\\[4pt] \dfrac1{2\sqrt{b^2-1}}\ln\left|\dfrac{y-\sqrt{b^2-1}}{y+\sqrt{b^2-1}}\right|, \text{ if }b>1, \end{cases}\tag3$$ $$I_2 = -\int\dfrac{y^2dy}{(y^2+1-b^2)^2} = \dfrac12\int y\cdot d\left(\dfrac{1}{y^2+1-b^2}\right) = \dfrac12\dfrac{y}{y^2+1-b^2} - \dfrac {I_1}2.\tag4$$

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Hint: $\sin x = \dfrac{\tan x}{\sec x}$ and $\sec^2x = 1 + \tan^2x$. $$\begin{align} \dfrac1{(\sin x+a\sec x)^2} &= \dfrac1{\left(\dfrac{\tan x}{\sec x}+a\sec x\right)^2} \\ &= \dfrac{\sec^2x}{(\tan x + a\sec^2x)^2} \\ &= \dfrac{\sec^2x}{\tan^2x + 2a\tan x\sec^2x + a^2\sec^4x} \\ &= \dfrac{\sec^2x}{\tan^2x + 2a\tan x(1 + \tan^2x)+ a^2(1 + \tan^2x)^2} \end{align}$$ Therefore, $$\int\dfrac1{(\sin x+a\sec x)^2}\,\mathrm dx = \int\dfrac{\sec^2x}{\tan^2x + 2a\tan x(1 + \tan^2x)+ a^2(1 + \tan^2x)^2}\,\mathrm dx$$ Substitute $u = \tan x$, $\implies\mathrm du = \sec^2x\,\mathrm dx$. Now, $$\begin{align}\int\dfrac{\sec^2x}{\tan^2x + 2a\tan x(1 + \tan^2x)+ a^2(1 + \tan^2x)^2}\,\mathrm dx &= \int \dfrac1{u^2 + 2au(1 + u^2) + a^2(1 + u^2)^2}\,\mathrm du\end{align}$$ From here onward, you would need to factor the denominator and perform partial fraction decomposition. Then, you will be able to apply linearity to integrate individual fractions. Finally, undo substitution to get the final result.

Edit # 1: $$\begin{align} u^2 + 2au(1 + u^2) + a^2(1 + u^2)^2 &\equiv u^2 + 2\cdot (u)\cdot\left(a(1 + u^2)\right) + \left(a(1 + u^2)\right)^2 \\ &= \left(u + a(1 + u^2)\right)^2 \end{align}$$ Factorising $u + a(1 + u^2)$: $$u + a(1 + u^2) \equiv au^2 + u + a$$ Completing square, $$\begin{align} au^2 + u + a &= au^2 + u + \dfrac1{4a} - \dfrac1{4a} + a \\ &= \left(\sqrt{a}u + \dfrac1{2\sqrt{a}}\right)^2 - \dfrac1{4a} + a \\ &= \dfrac{(2au + 1)^2}{4a} - \dfrac{1 - 4a^2}{4a} \\ &= \dfrac{(2au + 1)^2 - \left(\sqrt{1 - 4a^2}\right)^2}{4a} \\ &= \dfrac{\left(2au + 1 + \sqrt{1 - 4a^2}\right)\left(2au + 1 - \sqrt{1 - 4a^2}\right)}{4a} \end{align}$$ Therefore, $$\begin{align} \int \dfrac1{u^2 + 2au(1 + u^2) + a^2(1 + u^2)^2}\,\mathrm du &= \operatorname{\large\int} \dfrac1{\left(au^2 + u + a\right)^2}\,\mathrm du\\ &= \operatorname{\Large\int} \dfrac{1}{\left(\left(2au + 1 + \sqrt{1 - 4a^2}\right)\left(2au + 1 - \sqrt{1 - 4a^2}\right)/4a\right)^2}\,\mathrm du \\ &= \operatorname{\Large\int} \dfrac{16a^2}{\left(2au + 1 + \sqrt{1 - 4a^2}\right)^2\left(2au + 1 - \sqrt{1 - 4a^2}\right)^2}\,\mathrm du \end{align}$$

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    $\begingroup$ What is remarkable are the roots of the denominator of your last integrand and how simple becomes the result of the integration. $\to +1$ for the solution. $\endgroup$ – Claude Leibovici May 24 '18 at 5:08
  • $\begingroup$ Nice an4s would you like to explain How we can factorise it. $\endgroup$ – DXT May 24 '18 at 6:08
  • $\begingroup$ Factorise using : $$ (x+y)^2=x^2+2xy+y^2$$So you will have consequently your roots verifing : $$ u=a(1+u^2) $$ and solving it again give it your roots $\endgroup$ – Pagode May 26 '18 at 23:14
  • $\begingroup$ @DurgeshTiwari see edit. $\endgroup$ – an4s May 27 '18 at 4:44
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    $\begingroup$ And what about $a\ge\dfrac12$? $\endgroup$ – Yuri Negometyanov May 31 '18 at 14:23
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You can write $$ \frac{1}{(a+t+at^2)^2}=-\frac{1-2a^2+2at+2a^2t^2}{(-1+4a^2)(a+t+at^2)^2}+\frac{2a}{(-1+4a^2)(a+t+at^2)}. $$ Also is $$ -\int\frac{1-2a^2+2at+2a^2t^2}{(-1+4a^2)(a+t+at^2)^2}dt =\frac{1+2at}{(-1+4a^2)(a+t+at^2)} $$ and $$ \int \frac{2a}{(-1+4a^2)(a+t+at^2)}dt=\frac{4a}{(-1+4a^2)^{3/2}}\arctan\left(\frac{1+2at}{\sqrt{-1+4a^2}}\right). $$ Hence $$ \int \frac{1}{(a+t+a t^2)^2}dt= $$ $$ =\frac{1+2at}{(-1+4a^2)(a+t+a t^2)}+\frac{4a}{(-1+4 a^2)^{3/2}}\arctan\left(\frac{1+2at}{\sqrt{-1+4a^2}}\right). $$ Where we have used $$ \int\frac{1}{1+t^2}dt=\arctan(t)+c $$

In general holds $$ \int\frac{dt}{(t^2+at+b)^2}=\frac{2t+a}{(-a^2+4b)(t^2+at+b)}+\frac{4\arctan\left(\frac{2t+a}{\sqrt{-a^2+4b}}\right)}{(-a^2+4b)^{3/2}}. $$

QED

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