Question Number 115459 by john santu last updated on 26/Sep/20
$${I}=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\sqrt[{\mathrm{3}\:}]{\mathrm{1}+{x}^{\mathrm{3}} }}\:? \\ $$$${I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\mathrm{cos}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)\:{dx}\:=\:? \\ $$$$ \\ $$
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)](https://www.tinkutara.com/question/Q115461.png)
$${I}=\int\mathrm{cos}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)\:{dx}\: \\ $$$${I}=\int\:\left(\mathrm{cos}\:{x}\:\mathrm{cos}\:\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} \:{dx}\: \\ $$$${I}=\:\frac{\mathrm{1}}{\mathrm{4}}\int\:\left(\mathrm{cos}\:\mathrm{3}{x}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} \:{dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\int\:\left(\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{3}{x}\right)+\mathrm{2cos}\:\mathrm{3}{x}\mathrm{cos}\:{x}+\mathrm{cos}\:^{\mathrm{2}} {x}\right){dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\int\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{6}{x}+\mathrm{cos}\:\mathrm{4}{x}+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{x}+\frac{\mathrm{1}}{\mathrm{2}}\right){dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\int\left(\mathrm{1}+\mathrm{cos}\:\mathrm{6}{x}+\mathrm{cos}\:\mathrm{4}{x}+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{x}\right){dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}{x}+\frac{\mathrm{1}}{\mathrm{24}}\mathrm{sin}\:\mathrm{6}{x}+\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sin}\:\mathrm{4}{x}+\frac{\mathrm{3}}{\mathrm{16}}\mathrm{sin}\:\:\mathrm{2}{x}+{c} \\ $$$${now}\:{put}\:{border}\: \\ $$$${I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\mathrm{cos}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)\:{dx}\: \\ $$$${I}=\:\left[\frac{{x}}{\mathrm{4}}\:+\frac{\mathrm{sin}\:\mathrm{6}{x}}{\mathrm{24}}\:+\:\frac{\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{16}}\:+\:\frac{\mathrm{3}\:\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{16}}\:\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$${I}=\frac{\pi}{\mathrm{8}} \\ $$
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](https://www.tinkutara.com/question/Q115519.png)
$$\mathrm{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{3}} \right)^{\frac{\mathrm{4}}{\mathrm{3}}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }=−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\frac{−\mathrm{3}}{\mathrm{x}^{\mathrm{4}} }\mathrm{dx}}{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\right)^{\frac{\mathrm{4}}{\mathrm{3}}} } \\ $$$$=\left[\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}}−\mathrm{2} \\ $$
Answered by TANMAY PANACEA last updated on 26/Sep/20
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}} {xcos}^{\mathrm{2}} \left(\mathrm{2}{x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−{x}\right){cos}^{\mathrm{2}} \left(\pi−\mathrm{2}{x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} \mathrm{2}{x}\:{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}\right){cos}^{\mathrm{2}} \mathrm{2}{x}\:{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+{cos}\mathrm{4}{x}}{\mathrm{2}}{dx} \\ $$$$\mathrm{4}{I}=\left({x}+\frac{{sin}\mathrm{4}{x}}{\mathrm{4}}\right)\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\mathrm{4}{I}=\frac{\pi}{\mathrm{2}}\rightarrow\boldsymbol{{so}}\:\boldsymbol{{I}}=\frac{\pi}{\mathrm{8}} \\ $$