Jaya Legge

2021-06-25

Find a function f such that $f{}^{‴}\left(x\right)=\frac{1}{\sqrt{1+{\mathrm{sin}}^{2}x}}$. (This problem is supposed to be easy; don't misinterpret the word "find".)

Willie

If
Thus,
${\left(f{}^{″}\left(x\right)\right)}^{\prime }=\frac{1}{\sqrt{1+{\mathrm{sin}}^{2}\left(x\right)}}$
$f{}^{″}\left(x\right){=}^{\left(1\right)}\int \frac{1}{\sqrt{1+{\mathrm{sin}}^{2}+\left(x\right)}}dx$
The above equation, can be written as:
${\left({f}^{\prime }\left(x\right)\right)}^{\prime }=\int \frac{1}{\sqrt{1+{\mathrm{sin}}^{2}\left(x\right)}}dx$
So,
${f}^{\prime }\left(x\right){=}^{\left(1\right)}\int \int \frac{1}{\sqrt{1+{\mathrm{sin}}^{2}\left(x\right)}}dxdx$
Now, from (1) we get
$f\left(x\right)=\int \int \int \frac{1}{\sqrt{1+{\mathrm{sin}}^{2}\left(x\right)}}dxdxdx$

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