According to the standard integral results, it is known that $\int \csc^2 x \, dx = -\cot x$ since $\frac{d}{dx} \cot x = - \csc^2 x$. However, supposing that this is not known, is there a better method of evaluating $\int \csc^2 x\, dx$ ? I'm still a little new to integration.
Thanks!
$\endgroup$3 Answers
$\begingroup$Note that $$I = \int \csc^2 x \, dx = \int \frac{1}{\sin^2 x}\, dx = \int \frac{\sec^2 x}{\tan^2 x}\, dx$$
Can you see what will you get on substituting $u=\tan x$?
$\endgroup$ $\begingroup$$$\csc^2x=1+\cos x\cdot\dfrac{\cos x}{\sin^2x}$$
Integrate both sides, $$\int\csc^2x\ dx=\int\ dx+\int\cos x\cdot\dfrac{\cos x}{\sin^2x}\ dx$$
Now integrating by parts,
$$\int\cos x\cdot\dfrac{\cos x}{\sin^2x}\ dx=\cos x\int\dfrac{\cos x}{\sin^2x}\ dx-\int\left(\dfrac{d(\cos x)}{dx}\int\dfrac{\cos x}{\sin^2x}\ dx\right)dx$$
$$=\cos x\cdot\dfrac1{-\sin x}-\int\dfrac{\sin x}{\sin x}dx=?$$
$\endgroup$ $\begingroup$you can use that $$\csc^2(x)=\frac{1}{\sin^2(x)}=\frac{\sin^2(x)+\cos^2(x)}{\sin^2(x)}$$
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