Question Number 224124 by klipto last updated on 21/Aug/25
$$\int_{−\mathrm{2}} ^{\mathrm{2}} \frac{\boldsymbol{\mathrm{x}}^{\mathrm{3}} \boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}}{\:\sqrt{\mathrm{4}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }} \\ $$
Answered by klipto last updated on 21/Aug/25
![klipto−quanta♠ ∫_(−2) ^2 ((x^3 cos((x/2))+(1/2))/( (√(4−x^2 ))))dx if f(−x)=−f(x)−−−−function is odd and if f(x) is an odd function i.e ∫_(−k) ^k f(x)dx=0, note cos(−x)=cosx ∫_(−2) ^2 ((x^3 cos((x/2))+(1/2))/( (√(4−x^2 ))))dx=∫_(−2) ^2 (((x^3 cos((x/2)))/( (√(4−x^2 ))))+((1/2)/( (√(4−x^2 )))))dx testing: x^3 (odd),cos((x/2))(even),(√(4−x^2 )) (even) i.f(x)=((x^3 cos((x/2)))/( (√(4−x^2 )))),f(−x)=(((−x)^3 cos(−(x/2)))/( (√(4−(−x)^2 ))))=((−x^3 cos(x/2))/( (√(4−x^2 ))))=−f(x) i.e ∫_(−2) ^2 ((x^3 cos((x/2)))/( (√(4−x^2 ))))dx=0 ∫_(−2) ^2 ((1/2)/( (√(4−x^2 ))))dx =(1/2)∫_(−2) ^2 (1/( (√(4−x^2 ))))dx=(1/2)∫_(−2 ) ^2 (1/( (√(4(1−(x^2 /4))))))dx =(1/2)∫_(−2) ^2 (1/(2(√(1−((x/2))^2 ))))dx=(1/4)∫_(−2) ^2 (1/( (√(1−((x/2))^2 ))))dx u=(x/2),(du/dx)=(1/2),2du=dx (1/4)∫_(−2) ^2 ((2du)/( (√(1−u^2 ))))=(1/2)∫_(−2) ^2 (du/( (√(1−u^2 )))) =[((1/2)arcsin((x/2))]_(−2) ^2 =(1/2)[arcsin(1)−arcsin(−1)]=(1/2)(90^0 −(−90^0 )=90^0 =(𝛑/2)✓ ∴∫_(−2) ^2 ((x^3 cos((x/2))+(1/2))/( (√(4−x^2 ))))dx=(𝛑/2)≈1.570 klipto−quanta](https://www.tinkutara.com/question/Q224134.png)
$$ \\ $$$$\boldsymbol{\mathrm{klipto}}−\boldsymbol{\mathrm{quanta}}\spadesuit \\ $$$$\int_{−\mathrm{2}} ^{\mathrm{2}} \frac{\boldsymbol{\mathrm{x}}^{\mathrm{3}} \boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}}{\:\sqrt{\mathrm{4}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{f}}\left(−\boldsymbol{\mathrm{x}}\right)=−\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)−−−−\boldsymbol{\mathrm{function}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{odd}} \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{odd}}\:\boldsymbol{\mathrm{function}}\: \\ $$$$\boldsymbol{\mathrm{i}}.\boldsymbol{\mathrm{e}}\:\int_{−\mathrm{k}} ^{\mathrm{k}} \boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{dx}}=\mathrm{0},\: \\ $$$$\boldsymbol{\mathrm{note}}\:\boldsymbol{\mathrm{cos}}\left(−\boldsymbol{\mathrm{x}}\right)=\boldsymbol{\mathrm{cosx}} \\ $$$$\int_{−\mathrm{2}} ^{\mathrm{2}} \frac{\boldsymbol{\mathrm{x}}^{\mathrm{3}} \boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}}{\:\sqrt{\mathrm{4}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}\boldsymbol{\mathrm{dx}}=\int_{−\mathrm{2}} ^{\mathrm{2}} \left(\frac{\mathrm{x}^{\mathrm{3}} \boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{4}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}+\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\:\sqrt{\mathrm{4}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}\right)\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{testing}}:\:\boldsymbol{\mathrm{x}}^{\mathrm{3}} \left(\boldsymbol{\mathrm{odd}}\right),\boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\right)\left(\boldsymbol{\mathrm{even}}\right),\sqrt{\mathrm{4}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:\left(\boldsymbol{\mathrm{even}}\right) \\ $$$$\boldsymbol{\mathrm{i}}.\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\frac{\boldsymbol{\mathrm{x}}^{\mathrm{3}} \boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{4}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }},\boldsymbol{\mathrm{f}}\left(−\boldsymbol{\mathrm{x}}\right)=\frac{\left(−\boldsymbol{\mathrm{x}}\right)^{\mathrm{3}} \boldsymbol{\mathrm{cos}}\left(−\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{4}−\left(−\boldsymbol{\mathrm{x}}\right)^{\mathrm{2}} }}=\frac{−\boldsymbol{\mathrm{x}}^{\mathrm{3}} \boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}}{\:\sqrt{\mathrm{4}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}=−\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$$$\boldsymbol{\mathrm{i}}.\boldsymbol{\mathrm{e}} \\ $$$$\int_{−\mathrm{2}} ^{\mathrm{2}} \frac{\boldsymbol{\mathrm{x}}^{\mathrm{3}} \boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{4}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}\boldsymbol{\mathrm{dx}}=\mathrm{0} \\ $$$$\int_{−\mathrm{2}} ^{\mathrm{2}} \frac{\frac{\mathrm{1}}{\mathrm{2}}}{\:\sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }}\boldsymbol{\mathrm{dx}}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{2}} ^{\mathrm{2}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{4}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}\boldsymbol{\mathrm{dx}}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{2}\:} ^{\mathrm{2}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{4}\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4}}\right)}}\boldsymbol{\mathrm{dx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{2}} ^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−\left(\frac{\mathrm{x}}{\mathrm{2}}\right)^{\mathrm{2}} }}\boldsymbol{\mathrm{dx}}=\frac{\mathrm{1}}{\mathrm{4}}\int_{−\mathrm{2}} ^{\mathrm{2}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\left(\frac{\mathrm{x}}{\mathrm{2}}\right)^{\mathrm{2}} }}\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{u}}=\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}},\frac{\boldsymbol{\mathrm{du}}}{\boldsymbol{\mathrm{dx}}}=\frac{\mathrm{1}}{\mathrm{2}},\mathrm{2}\boldsymbol{\mathrm{du}}=\boldsymbol{\mathrm{dx}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int_{−\mathrm{2}} ^{\mathrm{2}} \frac{\mathrm{2du}}{\:\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{2}} ^{\mathrm{2}} \frac{\boldsymbol{\mathrm{du}}}{\:\sqrt{\mathrm{1}−\boldsymbol{\mathrm{u}}^{\mathrm{2}} }} \\ $$$$=\left[\left(\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{arcsin}}\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\right)\right]_{−\mathrm{2}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{arcsin}\left(\mathrm{1}\right)−\mathrm{arcsin}\left(−\mathrm{1}\right)\right]=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{90}^{\mathrm{0}} −\left(−\mathrm{90}^{\mathrm{0}} \right)=\mathrm{90}^{\mathrm{0}} =\frac{\boldsymbol{\pi}}{\mathrm{2}}\checkmark\right.\right. \\ $$$$\therefore\int_{−\mathrm{2}} ^{\mathrm{2}} \frac{\boldsymbol{\mathrm{x}}^{\mathrm{3}} \boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}}{\:\sqrt{\mathrm{4}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\approx\mathrm{1}.\mathrm{570} \\ $$$$\boldsymbol{\mathrm{klipto}}−\boldsymbol{\mathrm{quanta}} \\ $$
Commented by necx122 last updated on 23/Aug/25
$${I}\:{love}\:{your}\:{approach}\:{sir}\:{Klipto}. \\ $$
Commented by klipto last updated on 23/Aug/25
$$\mathrm{thanks}\:\mathrm{please}\:\mathrm{who}\:\mathrm{is}\:\mathrm{this}? \\ $$
Commented by necx122 last updated on 23/Aug/25
$${I}\:{love}\:{your}\:{approach}\:{sir}\:{Klipto}. \\ $$
Answered by Frix last updated on 22/Aug/25
![Or without overthinking: I=∫_(−a) ^a f(x)dx =^([t=−x]) −∫_a ^(−a) f(−t)dt=∫_(−a) ^a f(−x)dx ⇒ I=(1/2)∫_(−a) ^a (f(x)+f(−x))dx ⇒ ∫_(−2) ^2 (((1/2)+x^3 cos (x/2))/( (√(4−x^2 ))))dx=(1/2)∫_(−2) ^2 (dx/( (√(4−x^2 ))))=∫_0 ^2 (dx/( (√(4−x^2 ))))= =[sin^(−1) (x/2)]_0 ^2 =(π/2)](https://www.tinkutara.com/question/Q224141.png)
$$\mathrm{Or}\:\mathrm{without}\:\mathrm{overthinking}: \\ $$$${I}=\underset{−{a}} {\overset{{a}} {\int}}{f}\left({x}\right){dx}\:\overset{\left[{t}=−{x}\right]} {=}\:−\underset{{a}} {\overset{−{a}} {\int}}{f}\left(−{t}\right){dt}=\underset{−{a}} {\overset{{a}} {\int}}{f}\left(−{x}\right){dx} \\ $$$$\Rightarrow \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\underset{−{a}} {\overset{{a}} {\int}}\left({f}\left({x}\right)+{f}\left(−{x}\right)\right){dx} \\ $$$$\Rightarrow \\ $$$$\underset{−\mathrm{2}} {\overset{\mathrm{2}} {\int}}\frac{\frac{\mathrm{1}}{\mathrm{2}}+{x}^{\mathrm{3}} \mathrm{cos}\:\frac{{x}}{\mathrm{2}}}{\:\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\underset{−\mathrm{2}} {\overset{\mathrm{2}} {\int}}\frac{{dx}}{\:\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}=\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\frac{{dx}}{\:\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}= \\ $$$$=\left[\mathrm{sin}^{−\mathrm{1}} \:\frac{{x}}{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{2}} =\frac{\pi}{\mathrm{2}} \\ $$
Commented by klipto last updated on 22/Aug/25
$$\boldsymbol{\mathrm{frix}}\:\boldsymbol{\mathrm{bad}}\:\boldsymbol{\mathrm{man}} \\ $$