Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 115459 by john santu last updated on 26/Sep/20

I= ∫_0 ^1  (dx/((1+x^3 )((1+x^3 ))^(1/(3 )) )) ?  I=∫_0 ^(π/2)  cos^2 x cos^2 (2x) dx = ?

$${I}=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\sqrt[{\mathrm{3}\:}]{\mathrm{1}+{x}^{\mathrm{3}} }}\:? \\ $$$${I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\mathrm{cos}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)\:{dx}\:=\:? \\ $$$$ \\ $$

Answered by bemath last updated on 26/Sep/20

I=∫cos^2 x cos^2 (2x) dx   I=∫ (cos x cos (2x))^2  dx   I= (1/4)∫ (cos 3x+cos x)^2  dx  I=(1/4)∫ (cos^2 (3x)+2cos 3xcos x+cos^2 x)dx  I=(1/4)∫((1/2)+(1/2)cos 6x+cos 4x+(3/2)cos 2x+(1/2))dx  I=(1/4)∫(1+cos 6x+cos 4x+(3/2)cos 2x)dx  I=(1/4)x+(1/(24))sin 6x+(1/(16))sin 4x+(3/(16))sin  2x+c  now put border   I=∫_0 ^(π/2)  cos^2 x cos^2 (2x) dx   I= [(x/4) +((sin 6x)/(24)) + ((sin 4x)/(16)) + ((3 sin 2x)/(16)) ]_0 ^(π/2)   I=(π/8)

$${I}=\int\mathrm{cos}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)\:{dx}\: \\ $$$${I}=\int\:\left(\mathrm{cos}\:{x}\:\mathrm{cos}\:\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} \:{dx}\: \\ $$$${I}=\:\frac{\mathrm{1}}{\mathrm{4}}\int\:\left(\mathrm{cos}\:\mathrm{3}{x}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} \:{dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\int\:\left(\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{3}{x}\right)+\mathrm{2cos}\:\mathrm{3}{x}\mathrm{cos}\:{x}+\mathrm{cos}\:^{\mathrm{2}} {x}\right){dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\int\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{6}{x}+\mathrm{cos}\:\mathrm{4}{x}+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{x}+\frac{\mathrm{1}}{\mathrm{2}}\right){dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\int\left(\mathrm{1}+\mathrm{cos}\:\mathrm{6}{x}+\mathrm{cos}\:\mathrm{4}{x}+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{x}\right){dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}{x}+\frac{\mathrm{1}}{\mathrm{24}}\mathrm{sin}\:\mathrm{6}{x}+\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sin}\:\mathrm{4}{x}+\frac{\mathrm{3}}{\mathrm{16}}\mathrm{sin}\:\:\mathrm{2}{x}+{c} \\ $$$${now}\:{put}\:{border}\: \\ $$$${I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\mathrm{cos}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)\:{dx}\: \\ $$$${I}=\:\left[\frac{{x}}{\mathrm{4}}\:+\frac{\mathrm{sin}\:\mathrm{6}{x}}{\mathrm{24}}\:+\:\frac{\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{16}}\:+\:\frac{\mathrm{3}\:\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{16}}\:\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$${I}=\frac{\pi}{\mathrm{8}} \\ $$

Answered by Dwaipayan Shikari last updated on 26/Sep/20

I=∫_0 ^1 (dx/((1+x^3 )^(4/3) ))=∫_0 ^1 (dx/(x^4 (1+(1/x^3 ))^(4/3) ))=−(1/3)∫_0 ^1 ((((−3)/x^4 )dx)/((1+(1/x^3 ))^(4/3) ))  =[(1+(1/x^3 ))^(−(1/3)) ]_0 ^1 =(1/( (2)^(1/3) ))−2

$$\mathrm{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{3}} \right)^{\frac{\mathrm{4}}{\mathrm{3}}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }=−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\frac{−\mathrm{3}}{\mathrm{x}^{\mathrm{4}} }\mathrm{dx}}{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\right)^{\frac{\mathrm{4}}{\mathrm{3}}} } \\ $$$$=\left[\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}}−\mathrm{2} \\ $$

Answered by TANMAY PANACEA last updated on 26/Sep/20

I=∫_0 ^(π/2) cos^2 xcos^2 (2x)dx  =∫_0 ^(π/2) cos^2 ((π/2)−x)cos^2 (π−2x)dx  =∫_0 ^(π/2) sin^2 xcos^2 2x dx  2I=∫_0 ^(π/2) (cos^2 x+sin^2 x)cos^2 2x dx  2I=∫_0 ^(π/2) ((1+cos4x)/2)dx  4I=(x+((sin4x)/4))∣_0 ^(π/2)   4I=(π/2)→so I=(π/8)

$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}} {xcos}^{\mathrm{2}} \left(\mathrm{2}{x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−{x}\right){cos}^{\mathrm{2}} \left(\pi−\mathrm{2}{x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} \mathrm{2}{x}\:{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}\right){cos}^{\mathrm{2}} \mathrm{2}{x}\:{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+{cos}\mathrm{4}{x}}{\mathrm{2}}{dx} \\ $$$$\mathrm{4}{I}=\left({x}+\frac{{sin}\mathrm{4}{x}}{\mathrm{4}}\right)\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\mathrm{4}{I}=\frac{\pi}{\mathrm{2}}\rightarrow\boldsymbol{{so}}\:\boldsymbol{{I}}=\frac{\pi}{\mathrm{8}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com