Question

INTEGRATE THE FOLLOWING
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Answer 1

To Integrate :

$\star \: \sqrt{ \frac{1 + \sin \: x}{1 - \sin \: x} }$

Here,

• Rationalise the denominator ,we get :

$\implies \: \sqrt{\frac{1 + \sin \: x}{1 - \sin \: x} \times \frac{1 + \sin \: x}{1 + \sin \: x} } \\ \\ \implies \: \sqrt{ \frac{ {(1 + \sin \: x)}^{2} }{ {1}^{2} - { \sin}^{2}x }} \\ \\ \implies \: \sqrt{ \frac{ {(1 + \sin \: x)}^{2} }{ { \cos}^{2} x} } \\ \\ \implies \: \frac{1 + \sin \: x}{ \cos \:x} \\ \\ \implies \: \frac{1}{ \cos \: x} + \frac{ \sin \: x}{ \cos \: x} \\ \\ \implies \: \sec \: x + \tan \: x$

Now , integrate

$\implies \int \: (\sec \: x + \tan \: x) \: dx \\ \\ \implies \: \log( \sec \: x \: + \tan \: x) + \log( \sec \: x) + c$

Formula used:

$\star \bold{ \int \: \tan \: x \: dx \: = log( \sec \: x) + c} \\ \\ \star \bold{ \int \: \sec \: x \: dx = log( \sec \: x + \tan \: x) }$

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\sf{\int \sqrt{\dfrac{1+sinx}{1-sinx}}dx}$

♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\sf{\int \sqrt{\dfrac{1+\sin \left(x\right)}{1-\sin \left(x\right)}}dx=2\ln \left|-1+\tan \left(\dfrac{x}{2}\right)\right|-\ln \left(\sec ^2\left(\dfrac{x}{2}\right)\right)+C}}$

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\text { Apply } u-\text { substitution: } u=\tan \left(\dfrac{x}{2}\right)$

$=\int \dfrac{2\left(u+1\right)}{\left(u-1\right)\left(1+u^2\right)}du$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=2\cdot \int \dfrac{u+1}{\left(u-1\right)\left(1+u^2\right)}du$

$\text { Take the partial fraction of } \dfrac{u+1}{(u-1)\left(1+u^{2}\right)}: \quad \dfrac{1}{u-1}-\dfrac{u}{u^{2}+1}$

$=2\cdot \int \dfrac{1}{u-1}-\dfrac{u}{u^2+1}du$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=2\left(\int \dfrac{1}{u-1}du-\int \dfrac{u}{u^2+1}du\right)$

$\begin{array}{l}\int \dfrac{1}{u-1} d u=\ln |u-1| \\\\\int \dfrac{u}{u^{2}+1} d u=\dfrac{1}{2} \ln \left|u^{2}+1\right|\end{array}$

$=2\left(\ln \left|u-1\right|-\dfrac{1}{2}\ln \left|u^2+1\right|\right)$

$\mathrm{Substitute\:back}\:u=\tan \left(\dfrac{x}{2}\right)$

$=2\left(\ln \left|\tan \left(\dfrac{x}{2}\right)-1\right|-\dfrac{1}{2}\ln \left|\tan ^2\left(\dfrac{x}{2}\right)+1\right|\right)$

$\text { Simplify } 2\left(\ln \left|\tan \left(\dfrac{x}{2}\right)-1\right|-\frac{1}{2} \ln \left|\tan ^{2}\left(\dfrac{x}{2}\right)+1\right|\right): \quad 2 \ln \left|-1+\tan \left(\dfrac{x}{2}\right)\right|-\ln \left(\sec ^{2}\left(\dfrac{x}{2}\right)\right)$

$=2\ln \left|-1+\tan \left(\dfrac{x}{2}\right)\right|-\ln \left(\sec ^2\left(\dfrac{x}{2}\right)\right)$

$\mathrm{Add\:a\:constant\:to\:the\:solution}$

$\boxed{\sf{=2\ln \left|-1+\tan \left(\dfrac{x}{2}\right)\right|-\ln \left(\sec ^2\left(\dfrac{x}{2}\right)\right)+C}}$