Answered: "Computing ∫0πdx1+a2cos2(x)"

jisu61hbke

Answered question

2022-03-25

Computing

\int_{0}^{π} \frac{d x}{1 + a^{2} \cos^{2} (x)}

disolutoxz61

Beginner2022-03-26Added 12 answers

Bioche's rules say the right substitution should be

t = \tan x

Indeed, then

d x = \frac{d t}{1 + t^{2}}

so that

\int \frac{d x}{1 + a^{2} \cos^{2} (x)} = \int \frac{d t}{(1 + t^{2}) (1 + \frac{a^{2}}{1 + t^{2}})} = \int \frac{d t}{1 + a^{2} + t^{2}}

= \frac{1}{\sqrt{1 + a^{2}}}, \arctan \frac{t}{\sqrt{1 + a^{2}}}

Kamora Campbell

Beginner2022-03-27Added 13 answers

Here is another way. Consider the identity

\cos^{2} x = \frac{1 + \cos 2 x}{2}

and the integral is then transformed into

2 \int_{0}^{π} \frac{d x}{2 + a^{2} + a^{2} \cos 2 x}

and using substitution

2 x = t

we get

\int_{0}^{2 π} \frac{d t}{2 + a^{2} + a^{2} \cos t}

Since

\cos (2 π - t) = \cos t

we can see that the above integral is equal to

2 \int_{0}^{π} \frac{d t}{2 + a^{2} + a^{2} \cos t}

and using the standard formula

\int_{0}^{π} \frac{d x}{A + B \cos x} = \frac{π}{\sqrt{A^{2} - B^{2}}}

we can see that the integral in question is equal to

\frac{2 π}{\sqrt{{(2 + a^{2})}^{2} - a^{4}}} = \frac{π}{\sqrt{1 + a^{2}}}

The standard formula given above can be proved using the substitution

(A + B \cos x) (A - B \cos y) = A^{2} - B^{2}

and is valid for

A > | B |

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