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Guys please help me in evaluating this integral $$ \int \frac{1}{(\sin(x) + a \sec(x))^2}\,dx $$ I tried by converting $\sec(x)$ to $\cos(x)$ and by solving it became more complicated so guys please guide me further.

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    $\begingroup$ I have no idea how but take a look at it : [wolframalpha.com/input/… $\endgroup$ – mrtaurho Aug 3 '18 at 11:21
  • $\begingroup$ Have you tried Weierstrauss sub? $\endgroup$ – Henry Lee Aug 6 '18 at 21:05
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Hint: $$\dfrac1{(\sin x+a\sec x)^2}=\dfrac1{2(\sin x\cos x+a)^2}+\dfrac{\cos2x}{2(\sin x\cos x+a)^2}$$

The second part is elementary.

$$\dfrac1{(\sin x\cos x+a)^2}=\dfrac{\sec^2x(1+\tan^2x)}{(\tan x+a\tan^2x+a)^2}$$

Choose $\tan x=u$

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  • $\begingroup$ I didn't get what You did after saying elementary $\endgroup$ – Ritik Aug 3 '18 at 11:35
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    $\begingroup$ @Ritik, Divided numerator & denominator by $$\cos^4x$$ $\endgroup$ – lab bhattacharjee Aug 3 '18 at 11:37
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    $\begingroup$ For the second fraction, it is simpler to set $u=\sin2x$, so $du=2\cos2x$ and the integral becomes $\int\frac{1}{(u/2+a)^2}\,du$ $\endgroup$ – egreg Aug 3 '18 at 11:46
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    $\begingroup$ How you guys are so good at integration ?? $\endgroup$ – Ritik Aug 3 '18 at 11:50
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    $\begingroup$ @labbhattacharjee I don't think that dividing by $\cos^4x$ is the best way. $\endgroup$ – egreg Aug 3 '18 at 13:47

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