Evaluating $ \int \frac{1}{\sin x} dx $

Question

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 $-\cos x + C$ ?

Answer 1

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

Answer 2

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

Answer 3

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