Find $\int_{-\pi/2}^{\pi/2} \frac{\cos^3 x}{e^x+1} dx$

Question

Find the value of the following integral

$$ \int_{-\pi/2}^{\pi/2} \frac{\cos^3 x}{e^x+1} dx $$

I tried to use the twin integral with sin and get some result but nothing obvious came out. Also, Mathematica gives a very bizarre result! Any ideas how to solve analytically?

Does this answer your question? Evaluating $\int_{-a}^a \frac{f(x)+1}{b^x+1},\mathrm dx$ for a $y$-axis symmetrical function $f$ — Maximilian Janisch, Mar 09 '20 at 22:38

score 8 · Accepted Answer · answered May 12 '18 at 08:39

8

Hint: make the substitution $x\mapsto -x$ and add the two integrals. The result should be $2/3$.

answered May 12 '18 at 08:39

user246336

3,579

Thank for the hint. The sum of the two equal integrals is 2/3. Isn't the final result 1/3? – asd11 May 12 '18 at 08:46
Are you sure the sum is 2/3? According to WA, $\int_{-\pi/2}^{\pi/2}\cos^3(x),dx=4/3$. – user246336 May 12 '18 at 08:50
the integrand is $\dfrac{\cos^3(x)}{e^x+1}$, not just $\cos^3(x)$. – Mr Pie May 12 '18 at 08:53
@user477343 But the sum of $\dfrac{\cos^3(x)}{e^x+1}$ and $\dfrac{\cos^3(x)}{e^{-x}+1}$ is $\cos^3(x)$. – user246336 May 12 '18 at 08:55
@iqcd you are right. Thanks! – asd11 May 12 '18 at 08:55
Yes, indeed you are, as a matter of fact :) – Mr Pie May 12 '18 at 08:56

score 3 · Answer 2 · answered May 12 '18 at 09:48

Like Evaluate the integral $\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\,\mathrm dx$.,

if $\displaystyle I=\int_{-a}^a\dfrac{g(x)\ dx}{b^x+1}$

$\displaystyle I=\int_{-a}^a\dfrac{g(a-a-x)\ dx}{b^{a-a-x}+1}$

If $g(x)=g(-x)$ i.e., an even function,

$\displaystyle I=\int_{-a}^ag(x)\ dx$

Here $g(x)=\cos^3x=\dfrac{\cos3x+3\cos x}4$

Find $\int_{-\pi/2}^{\pi/2} \frac{\cos^3 x}{e^x+1} dx$

2 Answers2

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