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Evaluate $\displaystyle\int_{-3}^{3}\dfrac{\mathrm dx}{3+f(x)}$ if $f(x)\cdot f(-x)=9$

Attempt:

$I = \int_{-3}^{3}\dfrac{\mathrm dx}{3+f(x)}$

$I = \int_{-3}^3 \dfrac{\mathrm dx}{3+f(-x)}$

Then I added the two $I$s to see if it helps, but it didn't.

I even substituted $f(-x)= \dfrac{9}{f(x)} $ in second equation but that didn't help too.

Please provide a hint on how to solve it.

StubbornAtom
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Archer
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    There is another shortcut: since the value of integral is the same as long as $f) satisfies the condition, just assume $f$ is a constant function and you will get $f=3$. Then you would get the same answer too. – Szeto Aug 07 '18 at 07:34
  • Great! Thanks @Szeto. – Archer Aug 07 '18 at 09:26

3 Answers3

7

\begin{align*} \int_{-3}^3\frac{\mathrm dx}{3+f(x)}&=\int_{-3}^0\frac{\mathrm dx}{3+f(x)}+\int_0^3\frac{\mathrm dx}{3+f(x)}&&\\[4pt] &=\int_{3}^0\frac{-\mathrm dt}{3+f(-t)}+\int_0^3\frac{\mathrm dx}{3+f(x)}&&\text{making }x=-t\text{ when }-3\le x\le0\\[4pt] &=\int_0^3\frac{\mathrm dt}{3+\frac9{f(t)}}+\int_0^3\frac{\mathrm dx}{3+f(x)}&&\text{since }f(t)\cdot f(-t)=9\\[4pt] &=\int_0^3\left(\frac1{3+9/f(x)}+\frac1{3+f(x)}\right)\mathrm dx&&\\[4pt] &=\int_0^3\left(\frac{f(x)}{3\left[3+f(x)\right]}+\frac1{3+f(x)}\right)\mathrm dx&&\\[4pt] &=\int_0^3\frac{f(x)+3}{3\left[3+f(x)\right]}\mathrm dx&&\\[4pt] &=\frac13\int_0^3\mathrm dx&&\\[4pt] &=1 \end{align*}

an4s
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Ángel Mario Gallegos
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Like Evaluate the integral $\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\,\mathrm dx$. OR Integral $\int_{1}^{2011} \frac{\sqrt{x}}{\sqrt{2012 - x} + \sqrt{x}}dx$,

As $\displaystyle\int_a^bf(x)\,\mathrm dx=\int_a^bf(a+b-x)\,\mathrm dx$

So, if $\int_a^bf(x)\,\mathrm dx=I,$

$$2I=\int_a^b[f(x)+f(a+b-x)]\,\mathrm dx$$

Here $$2I=\int_{-3}^3\left(\dfrac1{3+f(x)}+\dfrac1{3+f(-x)}\right)\mathrm dx$$

$$=\int_{-3}^3\dfrac{3+f(x)+3+f(-x)}{9+3f(x)+3f(-x)+f(x)f(-x)}\mathrm dx$$

$$=\int_{-3}^3\dfrac{\mathrm dx}3$$

an4s
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lab bhattacharjee
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1

As $$\frac{f(x)+3}{f(x)+3}=\frac{\dfrac9{f(-x)}}{\dfrac9{f(-x)}+3}+\frac3{f(x)+3}=\frac3{f(-x)+3}+\frac3{f(x)+3},$$

by integration from $-3$ to $3$ $$6=6I.$$