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I am trying to solve the integral of $\frac{ \sqrt{\cos 2 x}}{\sin x}$. I converted this to $(\cot^2 x - 1)^{1/2}$ but after this I am stuck. I am not able to think of a suitable substitution. Any tips?

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    $\begingroup$ added LaTeX, guessing at what you mean. $\endgroup$ – GEdgar Aug 4 '14 at 12:34
  • $\begingroup$ Inroads can be made with the substitution $z=\sin \left( \frac{x}{2}\right) \cos \left( \frac{x}{2} \right) = \frac{1}{2} \sin (x)$ $\endgroup$ – ja72 Aug 4 '14 at 13:13
  • $\begingroup$ Hint : substitute $\sec x=t$ $\endgroup$ – pkwssis Aug 4 '14 at 13:21
  • $\begingroup$ math.stackexchange.com/questions/876175/… $\endgroup$ – lab bhattacharjee Aug 4 '14 at 15:09
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Write it as $$I=\displaystyle\int \dfrac{\sqrt {\cos 2x}}{\sin x}=\displaystyle\int \dfrac{\sqrt {2\cos^2 x-1}}{\sin x}$$

And substitute $\cos x=t$. You should get $$I=\displaystyle\int \dfrac{\sqrt{2t^2-1}}{t^2-1} dt$$

Now substitute $t=\dfrac{1}{u}$, to further get $$I=\displaystyle\int \dfrac{\sqrt{2-u^2}}{u(u^2-1)} du$$

Make a third substitution $2-u^2=z^2$. Hence, $$\begin{align} I &=\displaystyle\int\dfrac{z^2}{(z^2-1)(2-z^2)}dz \\ &=2\displaystyle\int \left(\dfrac{1}{2z^2-2}-\dfrac{1}{z^2-2}\right)dz \\ &=\sqrt 2 \tanh^{-1} {\left(\dfrac{z}{\sqrt 2}\right)}-\tanh^{-1}z+C. \end{align}$$

Substitute back to get your answer.

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i solve your problem its not that harder....

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  • $\begingroup$ +1 rather than typing Latex people here should support these answer images ;) $\endgroup$ – Archis Welankar Feb 28 '16 at 14:57
  • $\begingroup$ yeah it supports.. $\endgroup$ – zen Feb 28 '16 at 15:05
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$\bf{My\; Solution::}$ Given $$\displaystyle \int\frac{\sqrt{\cos 2x}}{\sin x}dx = \int\frac{\sqrt{\cos^2 x-\sin^2 x}}{(\cos x+\sin x)-(\cos x-\sin x)}dx$$

So Integral $$\displaystyle =\int \frac{\sqrt{(\cos x+\sin x)\cdot (\cos x-\sin x)}}{(\cos x+\sin x)-(\cos x-\sin x)}dx = \int\frac{\sqrt{\frac{\cos x+\sin x}{\cos x-\sin x}}}{\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)-1}dx$$

Now we Can write $$\displaystyle \frac{\cos x+\sin x}{\cos x-\sin x}=\tan\left(\frac{\pi}{4}+x\right)$$

so Integral Convert into $$\displaystyle \int \frac{\sqrt{\tan\left(\frac{\pi}{4}+x\right)}}{\tan\left(\frac{\pi}{4}+x\right)-1}dy$$

Now Let $$\displaystyle \left(\frac{\pi}{4}+x\right)=y\;,$$ Then $dx=dy$

So Integral Convert into $$\displaystyle \int\frac{\sqrt{\tan y}}{\tan y-1}dy\;,$$ Now Let $$\displaystyle \tan y=t^2\;,$$ Then $\sec^2 y dy =2tdt$

So Integral Convert into $$\displaystyle \int \frac{2t^2}{(t^2-1)\cdot (t^4+1)}dt = \int \left[\frac{(t^4+1)-(t^2-1)^2}{(t^4+1)\cdot (t^2-1)}\right]dt$$

So Integral $$\displaystyle = \int\frac{1}{t^2-1}dt-\int\frac{t^2-1}{t^4+1}dt = -\frac{1}{2}\ln \left|\frac{t-1}{t+1}\right|-\int\frac{\left(1-\frac{1}{t^2}\right)}{\left(t+\frac{1}{t}\right)^2-\left(\sqrt{2}\right)^2}dt$$ $$\displaystyle = -\frac{1}{2}\ln \left|\frac{t-1}{t+1}\right|-\frac{1}{2\sqrt{2}}\ln \left|\frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1}\right|+\mathcal{C}$$

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