Show that $$\int_0^{\pi/3}\frac{1}{1-\sin x}\,\mathrm dx=1+\sqrt{3}$$

Using the substitution $t=\tan\frac{1}{2}x$

$\frac{\mathrm dt}{\mathrm dx}=\frac{1}{2}\sec^2\frac{1}{2}x$

$\mathrm dx=2\cos^2\frac{1}{2}x\,\mathrm dt$

$=(2-2\sin^2\frac{1}{2}x)\,\mathrm dt$

How do you get this in the form of $t$ instead of $x$ using $\sin A=\dfrac{2t}{1+t^2}$ ?

$$=\int_0^{1/\sqrt3}\frac{2-2\sin^2\frac{1}{2}x}{1-\frac{2t}{1+t^2}}\mathrm dt\,??$$

  • $\begingroup$ Sorry that was a mistake I have just edited! $\endgroup$ – maxmitch Jun 11 '13 at 21:19
  • $\begingroup$ Still says $t=\tan\frac{1}{2x}$. Also, is it really $\int_0^{\frac{1}{2\pi}}$ or is it $\int_0^{\frac{1}{2}\pi}$? $\endgroup$ – Thomas Andrews Jun 11 '13 at 21:21
  • $\begingroup$ I edited your question to enhance the typesetting; you can read about it here. Spacing issues are addressed in this answer. $\endgroup$ – Lord_Farin Jun 11 '13 at 21:33
  • $\begingroup$ Whenever I see an integral with a simple numerator and a complicated denominator, my first instinct is to multiply top and bottom by something to reverse things. In this case, multiplying by $1+\sin x$ makes things a lot neater. Doesn't always work... $\endgroup$ – DJohnM Jun 11 '13 at 21:34
  • $\begingroup$ Thank you. Sorry i had to undo it. I just had written an extra bit and it got deleted in the process and I take so long at mathJax! $\endgroup$ – maxmitch Jun 11 '13 at 21:35

I think you mean

$$\int_0^{\pi/3} \frac{dx}{1-\sin{x}}$$

which may be accomplished using the substitution $t=\tan{\frac{x}{2}}$. Then

$$dt = \frac12 \sec^2{\frac{x}{2}} dx= \frac12 (1+\tan^2{\frac{x}{2}}) dx = \frac12(1+t^2) dx$$

so that $dx = 2 dt/(1+t^2)$ Also

$$t^2=\frac{1-\cos{x}}{1+\cos{x}} \implies \cos{x} = \frac{1-t^2}{1+t^2} \implies \sin{x} = \frac{2 t}{1+t^2} $$

so that

$$1-\sin{x} = \frac{1+t^2-2 t}{1+t^2} = \frac{(1-t)^2}{1+t^2}$$

Then the integral is

$$2 \int_0^{1/\sqrt{3}} \frac{dt}{1+t^2} \frac{1+t^2}{(1-t)^2} = 2 \left [ \frac{1}{1-t}\right]_0^{1/\sqrt{3}} = 1+\sqrt{3} $$

  • $\begingroup$ I don't quite understand how $\frac{1}{2}(1+\tan^2{\frac{x}{2}}) dx = \frac12(1+t^2) dx$ can you elaborate? $\endgroup$ – maxmitch Jun 11 '13 at 21:33
  • $\begingroup$ @maxmitch: remember the original substitution: $t=\tan{(x/2)}$. $\endgroup$ – Ron Gordon Jun 11 '13 at 21:35
  • $\begingroup$ Ok... I shouldn't be doing further maths. Haha thank you! $\endgroup$ – maxmitch Jun 11 '13 at 21:36
  • 1
    $\begingroup$ Dont say that buddy, go as far in math as you want! $\endgroup$ – zerosofthezeta Jun 11 '13 at 23:31

I'd just use the identity $\dfrac 1{1-\sin x}=\dfrac{1+\sin x}{1-\sin^2x}=\dfrac{1+\sin x}{\cos^2x}=\sec^2x+\sec x\tan x$.


The Weierstrass substitution sets $t = \tan \frac{x}{2}$ and it's possible that's what you mean to do. In that case, $$\frac{\mathrm{d}t}{\mathrm{d}x} = \frac{1}{2} \mathrm{sec}^2 \frac{x}{2} = \frac{1}{2} (1 + \tan^2 \frac{x}{2}) = \frac{1 + t^2}{2}$$ which gives you $$\mathrm{d}x = \frac{2}{1+t^2} \mathrm{d}t.$$


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