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Question Number 66325 by mathmax by abdo last updated on 12/Aug/19

$$calculate{\int }_0^{\frac{\pi }{4}}{cos}^{4}x{sin}^{2}xdx$$

Commented by Prithwish sen last updated on 13/Aug/19

$$\frac{1}{8}{\int }_0^{\frac{\pi }{4}}(\cos 2\mathrm{x}+1){\sin }^{2}2\mathrm{xdx}=\frac{1}{8}{\int }_0^{\frac{\pi }{4}}{\cos 2\mathrm{xsin}}^{2}2\mathrm{xdx}+\frac{1}{16}{\int }_0^{\frac{\pi }{4}}(1-\cos 4\mathrm{x})\mathrm{dx}$$$$={[\frac{{\sin }^{3}2\mathrm{x}}{48}+\frac{\mathrm{x}}{16}-\frac{\sin 4\mathrm{x}}{64}]}_0^{\frac{\pi }{4}}=\frac{1}{48}+\frac{\pi }{64}\mathrm{please}\mathrm{check}.$$

Commented by mathmax by abdo last updated on 13/Aug/19

$$letI={\int }_0^{\frac{\pi }{4}}{cos}^{4}x{sin}^{2}x\Rightarrow I={\int }_0^{\frac{\pi }{4}}{\left(\frac{1+cos\left(2x\right)}{2}\right)}^2\left(\frac{1-cos\left(2x\right)}{2}\right)dx$$$$=\frac{1}{8}{\int }_0^{\frac{\pi }{4}}(1+2cos\left(2x\right)+\frac{1+cos\left(4x\right)}{2})(1-cos\left(2x\right)dx$$$$=\frac{1}{16}{\int }_0^{\frac{\pi }{4}}(2+4cos\left(2x\right)+1+cos\left(4x\right))(1-cos\left(2x\right))dx$$$$=\frac{1}{16}{\int }_0^{\frac{\pi }{4}}(3+4cos\left(2x\right)+cos\left(4x\right))(1-cos\left(2x\right))dx$$$$16I={\int }_0^{\frac{\pi }{4}}(3+4cos\left(2x\right)+cos\left(4x\right)-3cos\left(2x\right)-4{cos}^2\left(2x\right)-cos\left(2x\right)cos\left(4x\right))dx$$$$=\frac{3\pi }{4}+{\int }_0^{\frac{\pi }{4}}cos\left(2x\right)dx+{\int }_0^{\frac{\pi }{4}}cos\left(4x\right)dx-4{\int }_0^{\frac{\pi }{4}}\frac{1+cos\left(4x\right)}{2}dx$$$$-{\int }_0^{\frac{\pi }{4}}cos\left(2x\right)cos\left(4x\right)dx$$$$=\frac{3\pi }{4}+\frac{1}{2}{\left[sin\left(2x\right)\right]}_0^{\frac{\pi }{4}}+\frac{1}{4}{\left[sin\left(4x\right)\right]}_0^{\frac{\pi }{4}}-\frac{\pi }{2}-\frac{1}{2}{\left[sin\left(4x\right)\right]}_0^{\frac{\pi }{4}}$$$$=\frac{3\pi }{4}+\frac{1}{2}-\frac{\pi }{2}=\frac{\pi }{4}+\frac{1}{2}$$

Commented by mathmax by abdo last updated on 13/Aug/19

$$erroratfinalline$$$$16I=\frac{3\pi }{4}+\frac{1}{2}-\frac{\pi }{2}-\frac{1}{2}{\int }_0^{\frac{\pi }{4}}(cos\left(6x\right)+cos\left(2x\right))dx$$$$but{\int }_0^{\frac{\pi }{4}}cos\left(6x\right)dx+{\int }_0^{\frac{\pi }{4}}cos\left(2x\right)dx=\frac{1}{6}{\left[sin\left(6x\right)\right]}_0^{\frac{\pi }{4}}+\frac{1}{2}{\left[sin\left(2x\right)\right]}_0^{\frac{\pi }{4}}$$$$=\frac{1}{6}sin\left(\frac{3\pi }{2}\right)+\frac{1}{2}=\frac{1}{2}-\frac{1}{6}\Rightarrow$$$$16I=\frac{\pi }{4}+\frac{1}{2}-\frac{1}{4}+\frac{1}{12}=\frac{\pi }{4}+\frac{1}{4}+\frac{1}{12}=\frac{\pi }{4}+\frac{1}{3}\Rightarrow$$$$I=\frac{\pi }{64}+\frac{1}{48}.$$

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