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

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.

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}

• 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. May 24, 2018 at 5:08
• Nice an4s would you like to explain How we can factorise it.
– DXT
May 24, 2018 at 6:08
• 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 May 26, 2018 at 23:14
• @DurgeshTiwari see edit.
– an4s
May 27, 2018 at 4:44
• And what about $a\ge\dfrac12$? May 31, 2018 at 14:23

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

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