# Finding f ( x ) in cos 2 </msup> &#x2061;<!-- ⁡ --> ( x

kolutastmr 2022-07-02 Answered
Finding $f\left(x\right)$ in ${\mathrm{cos}}^{2}\left(x\right)f\left(x\right)={x}^{2}-2{\int }_{1}^{x}\mathrm{sin}\left(t\right)\mathrm{cos}\left(t\right)f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t$
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## Answers (1)

Oliver Shepherd
Answered 2022-07-03 Author has 24 answers
Assuming your equation reads $f\left(x\right){\mathrm{cos}}^{2}x={x}^{2}-2{\int }_{1}^{x}\mathrm{sin}t\mathrm{cos}tf\left(t\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t$ then differentiating gives
$-2f\left(x\right)\mathrm{sin}x\mathrm{cos}x+{\mathrm{cos}}^{2}x{f}^{\prime }\left(x\right)=2x-2f\left(x\right)\mathrm{sin}x\mathrm{cos}x$
so upon simplification, this gives us ${f}^{\prime }\left(x\right){\mathrm{cos}}^{2}x=2x$ which is a differential equation you can solve
$f\left(x\right)=\int \frac{2x}{{\mathrm{cos}}^{2}x}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=\int 2x{\mathrm{sec}}^{2}x\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$
using IBP or some other technique you fancy. In particular, IBP gives
$\int 2x{\mathrm{sec}}^{2}x\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=2x\mathrm{tan}x-\int 2\mathrm{tan}x\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$
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