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Verify the identity

$$\sin x = \frac {2 \tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}}$$

Use this identity and the substitution $t = \tan\frac{x}{2}$ to evaluate the integral of $$ \int \frac{1}{\sin x} dx $$

My answer:

$$ A = \frac{x}{2} \Rightarrow \sin 2A= \frac{2\tan A}{1+\tan^2 A} = \frac{2\tan A}{\sec^2 A} = 2 \tan A\cos^2 A = 2 \sin A \cos A = \sin 2A $$

Since $x=A/2$, $\sin 2A = \sin x$

Let $t=\tan \frac{x}{2}$ $$ \int \frac{1}{\sin x} dx = \int \frac{2t}{1+t^2} dt $$

Let $u= 1+t^2$, $ du = 2t\,dt$

$$\int \frac{2t}{1+t^2} dt = \int \frac{2t}{u}\cdot\frac{1}{2t} du = \int \frac{1}{u} \,du = \ln u + C \\ = \ln(1+t^2) + C = \ln\left(1+\tan^2 \frac{x}{2} \right) + C $$

Then what do I do?

How do I show this is equal to $-\cos x + C$ ?

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    $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – user137731 Dec 9 '14 at 20:41
  • $\begingroup$ As it is, your post would require too much work (at least for me and, perhaps, others as well) to decypher what is written in it. Most probably that many won't even bother trying to decypher what is written. Read the other comment and learn the easy rules to properly write mathematics in this site. $\endgroup$ – Timbuc Dec 9 '14 at 20:44
  • $\begingroup$ This is not equal to $-\cos x+C$, beacuse you are integrating $1/\sin(x)$, not $\sin x$ (if I am correctly reading your question). $\endgroup$ – Przemysław Scherwentke Dec 9 '14 at 20:44
  • $\begingroup$ Why down vote? This post has shown a lot of work up vote from me. $\endgroup$ – jimjim Jan 5 '15 at 10:41
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Since $t=\tan \frac{x}{2}, \tan^{-1}t=\frac{x}{2}, x=2\tan^{-1}t$, and $dx=\frac{2}{1+t^2}dt$.

Then $\displaystyle\int\frac{1}{\sin x} dx=\int\frac{1+t^2}{2t}\cdot\frac{2}{1+t^2}dt=\int\frac{1}{t}dt=\ln|t|+C=\ln\left|\tan\frac{x}{2}\right|+C$

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$$I=\int\frac{dx}{\sin x}=\int\frac{1+\cos x}{\sin x}\frac{dx}{1+\cos x}=\int\frac{1+\cos x}{\sin x}d\left(\frac{\sin x}{1+\cos x}\right)$$ $$=\ln\left|\frac{\sin x}{1+\cos x}\right|+C=\ln\left|\tan {\frac{x}{2}}\right|+C$$

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$$\int\frac{1}{\sin (x)} dx=\int\frac{\sin(x)}{\sin^2(x)}dx=\int\frac{\sin(x)}{1-\cos^2(x)} $$ Use substitution $t=\cos(x) \to dt=\sin(x)dx\to dx=\frac{dt}{\sin(x)}$ $$\int \frac{\sin(x)}{1-t^2}*\frac{dt}{\sin(x)}=\int \frac{dt}{1-t^2}=arth(t)=arth(\cos(x))+C$$

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