jisu61hbke

2022-03-25

Computing ${\int }_{0}^{\pi }\frac{dx}{1+{a}^{2}{\mathrm{cos}}^{2}\left(x\right)}$

disolutoxz61

Bioche's rules say the right substitution should be $t=\mathrm{tan}x$
Indeed, then $dx=\frac{dt}{1+{t}^{2}}$ so that
$\int \frac{dx}{1+{a}^{2}{\mathrm{cos}}^{2}\left(x\right)}=\int \frac{dt}{\left(1+{t}^{2}\right)\left(1+\frac{{a}^{2}}{1+{t}^{2}}\right)}=\int \frac{dt}{1+{a}^{2}+{t}^{2}}$
$=\frac{1}{\sqrt{1+{a}^{2}}},\mathrm{arctan}\frac{t}{\sqrt{1+{a}^{2}}}$

Kamora Campbell

Here is another way. Consider the identity
${\mathrm{cos}}^{2}x=\frac{1+\mathrm{cos}2x}{2}$
and the integral is then transformed into
$2{\int }_{0}^{\pi }\frac{dx}{2+{a}^{2}+{a}^{2}\mathrm{cos}2x}$
and using substitution
$2x=t$
we get
${\int }_{0}^{2\pi }\frac{dt}{2+{a}^{2}+{a}^{2}\mathrm{cos}t}$
Since $\mathrm{cos}\left(2\pi -t\right)=\mathrm{cos}t$ we can see that the above integral is equal to
$2{\int }_{0}^{\pi }\frac{dt}{2+{a}^{2}+{a}^{2}\mathrm{cos}t}$
and using the standard formula
${\int }_{0}^{\pi }\frac{dx}{A+B\mathrm{cos}x}=\frac{\pi }{\sqrt{{A}^{2}-{B}^{2}}}$
we can see that the integral in question is equal to
$\frac{2\pi }{\sqrt{{\left(2+{a}^{2}\right)}^{2}-{a}^{4}}}=\frac{\pi }{\sqrt{1+{a}^{2}}}$
The standard formula given above can be proved using the substitution
$\left(A+B\mathrm{cos}x\right)\left(A-B\mathrm{cos}y\right)={A}^{2}-{B}^{2}$
and is valid for $A>|B|$

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