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Question Number 92420 by I want to learn more last updated on 06/May/20

Commented by mathmax by abdo last updated on 07/May/20

$$I=\int \frac{dx}{{cos}^{6}x+{sin}^{6}x}\Rightarrow I=\int \frac{dx}{{\left({cos}^{2}x\right)}^3+{\left({sin}^{2}x\right)}^3}$$$$=\int \frac{dx}{{({cos}^{2}x+{sin}^{2}x)}^3-3{cos}^{2}x{sin}^{2}x({cos}^{2}x+{sin}^{2}x)}$$$$=\int \frac{dx}{1-3{cos}^{2}x{sin}^{2}x}=\int \frac{dx}{1-3{\left(\frac{1}{2}sin\left(2x\right)\right)}^2}=\int \frac{dx}{1-3\times \frac{1}{4}{sin}^2\left(2x\right)}$$$$=4\int \frac{dx}{4-3{sin}^2\left(2x\right)}=4\int \frac{dx}{4-3\times \frac{1-cos\left(4x\right)}{2}}$$$$=8\int \frac{dx}{8-3+3cos\left(4x\right)}=8\int \frac{dx}{5+3cos\left(4x\right)}{=}_{4x=t}2\int \frac{dt}{5+3cost}$$$${=}_{tan\left(\frac{t}{2}\right)=u}2\int \frac{1}{5+3\times \frac{1-u^2}{1+u^2}}\times \frac{2du}{1+u^2}$$$$=4\int \frac{du}{5+5u^2+3-3u^2}=4\int \frac{du}{8+2u^2}=2\int \frac{du}{4+u^2}$$$${=}_{u=4\alpha }2\int \frac{4d\alpha }{4+4{\alpha }^2}=2\int \frac{d\alpha }{1+{\alpha }^2}=2arctan\left(\alpha \right)+C$$$$=2arctan\left(\frac{u}{4}\right)+C=2arctan\left(\frac{1}{4}tan\left(\frac{t}{2}\right)\right)+C$$$$=2arctan\left(\frac{1}{4}tan\left(2x\right)\right)+C$$

Commented by I want to learn more last updated on 12/May/20

$$\mathrm{Thanks}\mathrm{sir}$$

Answered by niroj last updated on 06/May/20

$$\int \frac{1}{{\cos }^6\mathrm{x}+{\sin }^6\mathrm{x}}\mathrm{dx}$$$$=\int \frac{1}{{\left({\sin }^2\mathrm{x}\right)}^3+{\left({\cos }^2\mathrm{x}\right)}^3}\mathrm{dx}$$$$=\int \frac{\mathrm{dx}}{({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x})({\sin }^4\mathrm{x}-{\sin }^2\mathrm{x}{\cos }^2\mathrm{x}+{\cos }^4\mathrm{x})}$$$$=\int \frac{1}{{({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x})}^2-{2\sin }^2\mathrm{x}{\cos }^2\mathrm{x}-{\sin }^2\mathrm{x}{\cos }^2\mathrm{x}}\mathrm{dx}$$$$=\int \frac{1}{1-{3\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}$$$$=\int \frac{{\sec }^2\mathrm{x}}{{\sec }^2\mathrm{x}-{3\tan }^2\mathrm{x}}\mathrm{dx}$$$$=\int \frac{{\sec }^2\mathrm{xdx}}{{\sec }^2\mathrm{x}-{\tan }^2\mathrm{x}-{2\tan }^2\mathrm{x}}$$$$=\int \frac{{\sec }^2\mathrm{x}\mathrm{dx}}{1-{2\tan }^2\mathrm{x}}$$$$\mathrm{Put},\tan \mathrm{x}=\mathrm{t}$$$${\sec }^2\mathrm{xdx}=\mathrm{dt}$$$$\int \frac{1}{1-{2\mathrm{t}}^2}\mathrm{dt}$$$$=\frac{1}{2}\int \frac{1}{\frac{1}{2}-{\mathrm{t}}^2}\mathrm{dt}$$$$=\frac{1}{2}\int \frac{1}{{\left(\sqrt{\frac{1}{2}}\right)}^2-{\left(\mathrm{t}\right)}^2}\mathrm{dt}$$$$=\frac{1}{2}.\frac{1}{2\sqrt{\frac{1}{2}}}\log \frac{\sqrt{\frac{1}{2}}+\mathrm{t}}{\sqrt{\frac{1}{2}}-\mathrm{t}}+\mathrm{C}$$$$=\frac{1}{2}\times \frac{1}{2\times \frac{1}{\sqrt{2}}}\log \frac{\frac{1}{\sqrt{2}}+\mathrm{t}}{\frac{1}{\sqrt{2}}-\mathrm{t}}+\mathrm{C}$$$$=\frac{\sqrt{2}}{4}\log \frac{1+\mathrm{t}\sqrt{2}}{1-\mathrm{t}\sqrt{2}}+\mathrm{C}$$$$=\frac{\sqrt{2}}{4}\times \frac{\sqrt{2}}{\sqrt{2}}\log \frac{1+\tan \mathrm{x}\sqrt{2}}{1-\tan \mathrm{x}\sqrt{2}}+\mathrm{C}$$$$=\frac{1}{2\sqrt{2}}\log \frac{1+\tan \mathrm{x}\sqrt{2}}{1-\tan \mathrm{x}\sqrt{2}}+\mathrm{C}//.$$

Commented by I want to learn more last updated on 06/May/20

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}$$

Commented by niroj last updated on 06/May/20

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Answered by niroj last updated on 07/May/20

$$\mathrm{Such}\mathrm{type}\mathrm{also}\mathrm{you}\mathrm{can}\mathrm{Apply}:$$$$\int \frac{1}{{\cos }^6\mathrm{x}+{\sin }^6\mathrm{x}}\mathrm{dx}$$$$\mathrm{dived}\mathrm{numerator}\mathrm{\&}\mathrm{denominator}\mathrm{by}{\cos }^6\mathrm{x}.$$$$=\int \frac{{\sec }^6\mathrm{x}\mathrm{dx}}{1+{\tan }^6\mathrm{x}}$$$$=\int \frac{{\sec }^4\mathrm{x}.{\sec }^2\mathrm{x}\mathrm{dx}}{{\left(1\right)}^3+{\left({\tan }^2\mathrm{x}\right)}^3}$$$$=\int \frac{{\left({\sec }^2\mathrm{x}\right)}^2{\sec }^2\mathrm{xdx}}{(1+{\tan }^2\mathrm{x})(1-{\tan }^2\mathrm{x}+{\tan }^4\mathrm{x})}$$$$=\int \frac{{(1+{\tan }^2\mathrm{x})}^2{\sec }^2\mathrm{xdx}}{(1+{\tan }^2\mathrm{x})(1-{\tan }^2\mathrm{x}+{\tan }^4\mathrm{x})}$$$$\mathrm{Put},\tan \mathrm{x}=\mathrm{t}$$$${\sec }^2\mathrm{xdx}=\mathrm{dt}$$$$\int \frac{{(1+{\mathrm{t}}^2)}^2\mathrm{dt}}{(1+{\mathrm{t}}^2)({\mathrm{t}}^4-{\mathrm{t}}^2+1)}$$$$=\int \frac{(1+{\mathrm{t}}^2)\mathrm{dt}}{{\mathrm{t}}^4-{\mathrm{t}}^2+1}$$$$=\int \frac{{\mathrm{t}}^2(1+\frac{1}{{\mathrm{t}}^2})}{{\mathrm{t}}^2({\mathrm{t}}^2-1+\frac{1}{{\mathrm{t}}^2})}\mathrm{dt}$$$$=\int \frac{(1+\frac{1}{{\mathrm{t}}^2})\mathrm{dt}}{{\mathrm{t}}^2-1+\frac{1}{{\mathrm{t}}^2}}=\int \frac{(1+\frac{1}{{\mathrm{t}}^2})\mathrm{dt}}{{(\mathrm{t}-\frac{1}{\mathrm{t}})}^2+2-1}$$$$\mathrm{Put},(\mathrm{t}-\frac{1}{\mathrm{t}})=\mathrm{m}$$$$(1+\frac{1}{{\mathrm{t}}^2})\mathrm{dt}=\mathrm{dm}$$$$\int \frac{1}{{\mathrm{m}}^2+1}\mathrm{dm}={\tan }^{-1}\mathrm{m}+\mathrm{C}$$$$={\tan }^{-1}(\mathrm{t}-\frac{1}{\mathrm{t}})+\mathrm{C}$$$$={\tan }^{-1}(\tan \mathrm{x}-\frac{1}{\tan \mathrm{x}})+\mathrm{C}$$$$={\tan }^{-1}(\tan \mathrm{x}-\cot \mathrm{x})+\mathrm{C}//.$$$$$$

Commented by I want to learn more last updated on 07/May/20

$$\mathrm{Thanks}\mathrm{sir},\mathrm{i}\mathrm{appreciate}\mathrm{your}\mathrm{time}$$

Commented by niroj last updated on 07/May/20

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