Question

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

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

2021-06-26
If $$\displaystyle{f}'{\left({x}\right)}={g{{\left({x}\right)}}}\ {t}{h}{e}{n}\ {f{{\left({x}\right)}}}=\int{g{{\left({x}\right)}}}{\left.{d}{x}\right.}$$
Thus,
$$\displaystyle{\left({f}{''}{\left({x}\right)}\right)}'={\frac{{{1}}}{{\sqrt{{{1}+{{\sin}^{{{2}}}{\left({x}\right)}}}}}}}$$
$$\displaystyle{f}{''}{\left({x}\right)}=^{{{\left({1}\right)}}}\int{\frac{{{1}}}{{\sqrt{{{1}+{{\sin}^{{{2}}}+}{\left({x}\right)}}}}}}{\left.{d}{x}\right.}$$
The above equation, can be written as:
$$\displaystyle{\left({f}'{\left({x}\right)}\right)}'=\int{\frac{{{1}}}{{\sqrt{{{1}+{{\sin}^{{{2}}}{\left({x}\right)}}}}}}}{\left.{d}{x}\right.}$$
So,
$$\displaystyle{f}'{\left({x}\right)}=^{{{\left({1}\right)}}}\int\int{\frac{{{1}}}{{\sqrt{{{1}+{{\sin}^{{{2}}}{\left({x}\right)}}}}}}}{\left.{d}{x}\right.}{\left.{d}{x}\right.}$$
Now, from (1) we get
$$\displaystyle{f{{\left({x}\right)}}}=\int\int\int{\frac{{{1}}}{{\sqrt{{{1}+{{\sin}^{{{2}}}{\left({x}\right)}}}}}}}{\left.{d}{x}\right.}{\left.{d}{x}\right.}{\left.{d}{x}\right.}$$