Question

$\int \frac{1}{\sqrt{1-\sin x}}dx=$ for $x\in \left(\frac{\pi }{4}\frac{\pi }{2}\right)$

• A. $\sqrt{2}\tan (\frac{3\pi }{8}+\frac{x}{4})+c$
• B. $\sqrt{2}\cot (\frac{3\pi }{8}+\frac{x}{4})+c$
• C. $\sqrt{2}\log |\cot (\frac{3\pi }{8}+\frac{x}{4})|+c$
• D. $\sqrt{2}\log |\cot (\frac{\pi }{8}-\frac{x}{4})|+c$

Answer 1

The correct option is D $\sqrt{2}\log |\cot (\frac{\pi }{8}-\frac{x}{4})|+c$

$\int \frac{1}{\sqrt{1-sinx}}dx=\int \frac{dx}{(si{n}^x{/}_2-co{s}^x{/}_2)}$
$\frac{1}{\sqrt{2}}\int \frac{dx}{{(}^1{/}_{\sqrt{2}})(si{n}^x{/}_2-co{s}^x{/}_2)}$
$\frac{1}{\sqrt{2}}\int \frac{dx}{sin{(}^{\pi }{/}_4{-}^x{/}_2)}$
$=\frac{1}{\sqrt{2}}\int cosec{(}^{\pi }{/}_4{-}^x{/}_2)dx$
$=\frac{2}{\sqrt{2}}\left[log\left|cosec{(}^{\pi }{/}_4{-}^x{/}_2\right)-cot{(}^{\pi }{/}_4{-}^x{/}_2\left)\right|\right]+c$
$=\sqrt{2}\left[log\left[cot{(}^{\pi }{/}_8{-}^x{/}_4\right)]\right]+c$
$=\sqrt{2}log\left|cot{(}^{\pi }{/}_8{-}^x{/}_4\right)|+c$