Finding $f(x)$ in ${\mathrm{cos}}^{2}(x)f(x)={x}^{2}-2{\int}_{1}^{x}\mathrm{sin}(t)\mathrm{cos}(t)f(t)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t$

kolutastmr
2022-07-02
Answered

Finding $f(x)$ in ${\mathrm{cos}}^{2}(x)f(x)={x}^{2}-2{\int}_{1}^{x}\mathrm{sin}(t)\mathrm{cos}(t)f(t)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t$

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Oliver Shepherd

Answered 2022-07-03
Author has **24** answers

Assuming your equation reads $f(x){\mathrm{cos}}^{2}x={x}^{2}-2{\int}_{1}^{x}\mathrm{sin}t\mathrm{cos}tf(t)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t$ then differentiating gives

$-2f(x)\mathrm{sin}x\mathrm{cos}x+{\mathrm{cos}}^{2}x{f}^{\prime}(x)=2x-2f(x)\mathrm{sin}x\mathrm{cos}x$

so upon simplification, this gives us ${f}^{\prime}(x){\mathrm{cos}}^{2}x=2x$ which is a differential equation you can solve

$f(x)=\int \frac{2x}{{\mathrm{cos}}^{2}x}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=\int 2x{\mathrm{sec}}^{2}x\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$

using IBP or some other technique you fancy. In particular, IBP gives

$\int 2x{\mathrm{sec}}^{2}x\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=2x\mathrm{tan}x-\int 2\mathrm{tan}x\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$

$-2f(x)\mathrm{sin}x\mathrm{cos}x+{\mathrm{cos}}^{2}x{f}^{\prime}(x)=2x-2f(x)\mathrm{sin}x\mathrm{cos}x$

so upon simplification, this gives us ${f}^{\prime}(x){\mathrm{cos}}^{2}x=2x$ which is a differential equation you can solve

$f(x)=\int \frac{2x}{{\mathrm{cos}}^{2}x}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=\int 2x{\mathrm{sec}}^{2}x\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$

using IBP or some other technique you fancy. In particular, IBP gives

$\int 2x{\mathrm{sec}}^{2}x\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=2x\mathrm{tan}x-\int 2\mathrm{tan}x\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$

asked 2022-06-06

Using the FTOC to find the derivative of the integral

${\int}_{{x}^{2}}^{5}(4x+2)\phantom{\rule{thickmathspace}{0ex}}dx.$

${\int}_{{x}^{2}}^{5}(4x+2)\phantom{\rule{thickmathspace}{0ex}}dx.$

asked 2022-06-24

We know if $g$ is continuous on $(a,b)$ and $F(x)={\int}_{a}^{x}g(t)dt$, then

${F}^{\prime}(x)=g(x)$

But, how about if we have

$F(x)={\int}_{a}^{h(x)}g(t)dt$

What should ${F}^{\prime}(x)$ be?? can we still apply fundamental theorem of calculus?

${F}^{\prime}(x)=g(x)$

But, how about if we have

$F(x)={\int}_{a}^{h(x)}g(t)dt$

What should ${F}^{\prime}(x)$ be?? can we still apply fundamental theorem of calculus?

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How do you use the Fundamental Theorem of Calculus to find the derivative of $\int \frac{1}{1+{t}^{2}}dt$ from $x$ to $5$?

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Can use FToC to evaluate $\underset{x\to \mathrm{\infty}}{lim}\frac{{\int}_{0}^{x}\phantom{\rule{mediummathspace}{0ex}}f\left(t\right)dt}{{x}^{2}}$?

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How does the first fundamental theorem of calculus guarantee the existence of antiderivatives of functions?

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Find the derivative of ${\int}_{3}^{x}{\mathrm{sin}}^{3}tdt$

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Let $f:[a,b]\to \mathbb{R}$ be monotone increasing, $F:[a,b]\to \mathbb{R}$ such that $F(x):={\int}_{[a,x]}f$ and ${x}_{0}\in [a,b]$. Then $F$ is differentiable at ${x}_{0}$ if and only if $f$ is continuous at ${x}_{0}$