If $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $. Find $f(x)$

Question

Problem: If $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $. Find $f(x)$

Solution: $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $

Differenting both sides,we get

$ f(x) \sin{x} \cos{x} = \frac {f'(x)}{2(b^2 - a^2)f(x)} $

Am I doing right ?

André Nicolas · Accepted Answer · 2013-07-24T15:50:02.297

7

It is a good start. For simplicity write $y$ for $f(x)$. We can rewrite the result you got as $$\frac{y'}{y^2}=2(b^2-a^2)\sin x\cos x.$$ Integrate both sides. It may be handy to note that $2\sin x\cos x=\sin(2x)$. Or not, since it is clear that $2\sin x\cos x$ is the derivative of $\sin^2 x$.

edited Jul 24 '13 at 15:50

answered Jul 24 '13 at 15:43

André Nicolas

507,029

Thanks.Now it becomes a question of differential equation – rst Jul 24 '13 at 15:51
@rst A really simple differential equation... – Thomas Andrews Jul 24 '13 at 15:51
Yes, but it is a separable DE, so it comes down to a couple of integrals both of which are familiar to you. – André Nicolas Jul 24 '13 at 15:52

score 5 · Answer 2 · edited Jul 24 '13 at 15:49

5

Yes and to complete: we have

$$\int f'(x)(f(x))^{-2}\;dx=C\int\sin(2x)\;dx$$ where $C=b^2-a^2$ so $$-\frac{1}{f(x)}=-\frac{C}{2}\cos(2x)+C'$$ and you can take $f(x)$ from it.

edited Jul 24 '13 at 15:49

Thomas Andrews

177,126

answered Jul 24 '13 at 15:45

If $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $. Find $f(x)$

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