8 votes

After integrating $\csc^2(x)\cot(x)$ why do I get $\cot^{2}(x) = \csc^{2}(x)$?

They differ by a constant: \begin{align*} \cot^{2}(x) = \frac{\cos^{2}(x)}{\sin^{2}(x)} = \frac{\cos^{2}(x) + \sin^{2}(x)}{\sin^{2}(x)} - 1 = \frac{1}{\sin^{2}(x)} - 1 = \csc^{2}(x) - 1 \end{align*}
Átila Correia's user avatar
4 votes

The fastest solution for $\int \sqrt{x^2+1}\,dx$

Another possible approach based on the integration by parts method: \begin{align*} \int\sqrt{x^{2} + 1}\mathrm{d}x & = x\sqrt{x^{2} + 1} - \int\frac{x^{2}}{\sqrt{x^{2} + 1}}\mathrm{d}x\\\\ & = ...
Átila Correia's user avatar
2 votes

After integrating $\csc^2(x)\cot(x)$ why do I get $\cot^{2}(x) = \csc^{2}(x)$?

This is because of the pesky $+C$ integration constant you must add at the end of every indefinite integral. There's a trig identity similar to $\sin^2\theta+\cos^2\theta=1$ that relates $\csc x$ to $\...
Frank W's user avatar
  • 5,932
1 vote

Solve integral of irrational function $\int \:\frac{2x^2-x+1}{\sqrt{2x^2+x-1}}dx$

Start with $$2x^2+x-1=(\sqrt2x+\frac 1{2\sqrt2})^2-\frac 98$$ so $$\int \frac{2x^2-x+1}{\sqrt{2x^2+x-1}}dx=\int \frac{2x^2-x+1}{\sqrt{(\sqrt2x+\frac 1{2\sqrt2})^2-\frac 98}}dx=\\2\sqrt2\int \frac{2x^2-...
Khosrotash's user avatar

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