Find the range of the function $f(x)=\mathrm{cos}(x)[\mathrm{sin}(x)+\sqrt{{\mathrm{sin}}^{2}(x)+\frac{1}{2}}]$

Brooke Richard
2022-11-21
Answered

Find the range of the function $f(x)=\mathrm{cos}(x)[\mathrm{sin}(x)+\sqrt{{\mathrm{sin}}^{2}(x)+\frac{1}{2}}]$

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boursecasa2je

Answered 2022-11-22
Author has **15** answers

$f(x)-\mathrm{cos}x\mathrm{sin}x=\mathrm{cos}x\sqrt{\frac{1}{2}+{\mathrm{sin}}^{2}x}\Rightarrow f(x{)}^{2}-2f(x)\mathrm{cos}x\mathrm{sin}x=\frac{1}{2}{\mathrm{cos}}^{2}x\Rightarrow \phantom{\rule{0ex}{0ex}}4f(x)\mathrm{cos}x\mathrm{sin}x=2f(x{)}^{2}-{\mathrm{cos}}^{2}x\Rightarrow 16f(x{)}^{2}{\mathrm{cos}}^{2}x(1-{\mathrm{cos}}^{2}x)=(2f(x{)}^{2}-{\mathrm{cos}}^{2}x{)}^{2}$

$f(x{)}^{2}=y,\text{}{\mathrm{cos}}^{2}x=t:$

$16yt(1-t)=(2y-t{)}^{2}\Rightarrow 4{y}^{2}+(16{t}^{2}-20t)y+{t}^{2}=0\Rightarrow \phantom{\rule{0ex}{0ex}}8y{y}^{\prime}+(16{t}^{2}-20t){y}^{\prime}+(32t-20)y+2t=0$

Here ${y}^{\prime}=\frac{dy}{dt}$. In critical points $\frac{df(x)}{dx}=0\Rightarrow \frac{dy}{dt}=0$

$(32t-20)y+2t=0\Rightarrow y=-\frac{t}{16t-10}\Rightarrow \frac{4{t}^{2}}{(16t-10{)}^{2}}-\frac{t(16{t}^{2}-20t)}{16t-10}+{t}^{2}=0\Rightarrow \phantom{\rule{0ex}{0ex}}4{t}^{2}-t(16{t}^{2}-20t)(16t-10)+{t}^{2}(16t-10{)}^{2}=0\Rightarrow 32{t}^{2}(5t-3)=0$

$t=0$ gives minimum $y=0$. $t=\frac{3}{5}$ gives maximum $y=\frac{3}{2}$

Minimum $f(x)=-\sqrt{\frac{3}{2}}$ is obtained at $\mathrm{cos}x=-\sqrt{\frac{3}{5}}$, $\mathrm{sin}x=\sqrt{\frac{2}{5}}$

Maximum $f(x)=\sqrt{\frac{3}{2}}$ is obtained at $\mathrm{cos}x=\sqrt{\frac{3}{5}}$, $\mathrm{sin}x=\sqrt{\frac{2}{5}}$

$f(x{)}^{2}=y,\text{}{\mathrm{cos}}^{2}x=t:$

$16yt(1-t)=(2y-t{)}^{2}\Rightarrow 4{y}^{2}+(16{t}^{2}-20t)y+{t}^{2}=0\Rightarrow \phantom{\rule{0ex}{0ex}}8y{y}^{\prime}+(16{t}^{2}-20t){y}^{\prime}+(32t-20)y+2t=0$

Here ${y}^{\prime}=\frac{dy}{dt}$. In critical points $\frac{df(x)}{dx}=0\Rightarrow \frac{dy}{dt}=0$

$(32t-20)y+2t=0\Rightarrow y=-\frac{t}{16t-10}\Rightarrow \frac{4{t}^{2}}{(16t-10{)}^{2}}-\frac{t(16{t}^{2}-20t)}{16t-10}+{t}^{2}=0\Rightarrow \phantom{\rule{0ex}{0ex}}4{t}^{2}-t(16{t}^{2}-20t)(16t-10)+{t}^{2}(16t-10{)}^{2}=0\Rightarrow 32{t}^{2}(5t-3)=0$

$t=0$ gives minimum $y=0$. $t=\frac{3}{5}$ gives maximum $y=\frac{3}{2}$

Minimum $f(x)=-\sqrt{\frac{3}{2}}$ is obtained at $\mathrm{cos}x=-\sqrt{\frac{3}{5}}$, $\mathrm{sin}x=\sqrt{\frac{2}{5}}$

Maximum $f(x)=\sqrt{\frac{3}{2}}$ is obtained at $\mathrm{cos}x=\sqrt{\frac{3}{5}}$, $\mathrm{sin}x=\sqrt{\frac{2}{5}}$

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$\int {\mathrm{cos}}^{2}x\mathrm{sin}2x\text{}dx$

I solved it using the identity ${\mathrm{cos}}^{2}x=\frac{1}{2}(1+\mathrm{cos}(2x))$ and the substitution $u=2x$

$\int {\mathrm{cos}}^{2}x\mathrm{sin}2x\text{}dx=\frac{1}{2}\int (1+\mathrm{cos}2x)\mathrm{sin}2x\text{}dx$

$\begin{array}{}\text{(}u=2x\text{)}& \frac{1}{4}\int (\mathrm{sin}u+\mathrm{sin}u\mathrm{cos}u)\text{}du\end{array}$

$\begin{array}{}\text{(}t=\mathrm{sin}u\text{)}& =\frac{1}{4}(\mathrm{cos}u+\int t\text{}dt)\end{array}$

$=\frac{\mathrm{cos}2x}{4}+\frac{{\mathrm{sin}}^{2}2x}{8}+C$

I can't find an error in my procedure.

$\int {\mathrm{cos}}^{2}x\mathrm{sin}2x\text{}dx$

I solved it using the identity ${\mathrm{cos}}^{2}x=\frac{1}{2}(1+\mathrm{cos}(2x))$ and the substitution $u=2x$

$\int {\mathrm{cos}}^{2}x\mathrm{sin}2x\text{}dx=\frac{1}{2}\int (1+\mathrm{cos}2x)\mathrm{sin}2x\text{}dx$

$\begin{array}{}\text{(}u=2x\text{)}& \frac{1}{4}\int (\mathrm{sin}u+\mathrm{sin}u\mathrm{cos}u)\text{}du\end{array}$

$\begin{array}{}\text{(}t=\mathrm{sin}u\text{)}& =\frac{1}{4}(\mathrm{cos}u+\int t\text{}dt)\end{array}$

$=\frac{\mathrm{cos}2x}{4}+\frac{{\mathrm{sin}}^{2}2x}{8}+C$

I can't find an error in my procedure.

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