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Question Number 214793 by vijaysahu last updated on 19/Dec/24

Commented by Tinku Tara last updated on 20/Dec/24

$$Q2putu=\frac{\pi }{2}-xtoget$$$$I={\int }_0^{\pi /4}\frac{{cos}^{4}x}{{sin}^{4}x+{cos}^{4}x}dx$$$$2I={\int }_0^{\pi /4}1dx=\frac{\pi }{4}$$

Commented by Tinku Tara last updated on 20/Dec/24

$$Q8isstraightforwardformula$$

Answered by MathematicalUser2357 last updated on 20/Dec/24

$$Q1\int \frac{1}{\sqrt{\tan x}+1}dx=\int {\sec }^{2}x\cdot \frac{1}{(\sqrt{\tan x}+1)({\tan }^{2}x+1)}dx$$$$\stackrel{u=\tan x\Rightarrow du={\sec }^{2}xdx}{\Rightarrow }\int \frac{1}{(\sqrt{u}+1)(u^2+1)}du$$$$\stackrel{v=\sqrt{u}\Rightarrow du=\frac{1}{2\sqrt{u}}du}{\Rightarrow }2\int \frac{v}{(v+1)(v^4+1)}dv$$$$\mathbf{Now}{\mathbf{we}}^{\prime }\mathbf{re}\mathbf{integrating}\mathbf{the}\mathbf{red}\mathbf{side}$$$$\int \frac{v}{(v+1)(v^4+1)}dv=\int (\frac{v^3-v^2+v+1}{2(v^4+1)}-\frac{1}{2(v+1)})dv$$$$=\frac{1}{2}\int \frac{v^3-v^2+v+1}{v^4+1}dv-\frac{1}{2}\int \frac{1}{v+1}dv$$$$\mathbf{Now}{\mathbf{we}}^{\prime }\mathbf{re}\mathbf{integrating}\mathbf{the}\mathbf{blue}\mathbf{side}$$$$\int \frac{v^3-v^2+v+1}{v^4+1}dv=\int \frac{v^3-v^2+v+1}{(v^2-\sqrt{2}v+1)(u^2+\sqrt{2}v+1)}dv$$$$=\int (\frac{(\sqrt{2}+2)v+\sqrt{2}}{2^{\frac{3}{2}}(v^2+\sqrt{2}v+1)}+\frac{(\sqrt{2}-2)v+\sqrt{2}}{2^{\frac{3}{2}}(v^2-\sqrt{2}+1)})dv$$$$=\frac{1}{2^{\frac{3}{2}}}\int \frac{(\sqrt{2}+2)v+\sqrt{2}}{v^2+\sqrt{2}v+1}dv+\frac{1}{2^{\frac{3}{2}}}\int \frac{(\sqrt{2}-2)v+\sqrt{2}}{v^2-\sqrt{2}v+1}dv$$$$\mathbf{Now}{\mathbf{we}}^{\prime }\mathbf{re}\mathbf{integrating}\mathbf{the}\mathbf{brown}\mathbf{side}$$$$\int \frac{(\sqrt{2}+2)v+\sqrt{2}}{v^2+\sqrt{2}v+1}dv=\int (\frac{(\sqrt{2}+2)(2v+\sqrt{2})}{2(v^2+\sqrt{2}v+1)}+\frac{\sqrt{2}-\frac{\sqrt{2}+2}{\sqrt{2}}}{v^2+\sqrt{2}v+1})dv$$$$=\frac{\sqrt{2}+2}{2}\int \frac{2v+\sqrt{2}}{v^2+\sqrt{2}v+1}dv+(\sqrt{2}-\frac{\sqrt{2}+2}{\sqrt{2}})\int \frac{1}{v^2+\sqrt{2}v+1}dv$$$$\mathcal{H}\mathrm{oly}...\mathrm{god}...\mathrm{what}\mathrm{a}\mathrm{long}\mathrm{integral}\mathrm{expression}$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\int \frac{2v+\sqrt{2}}{v^2+\sqrt{2}v+1}dv\stackrel{w=v^2+\sqrt{2}v+1\Rightarrow dv=(2v+\sqrt{2})dv}{\Rightarrow }\int \frac{1}{w}dw=\ln w=\ln (v^2+\sqrt{2}v+1)$$$$\int \frac{1}{v^2+\sqrt{2}v+1}dv=\int \frac{1}{{(v+\frac{1}{\sqrt{2}})}^2+\frac{1}{2}}dv$$$$\stackrel{w=\sqrt{2}v+1\Rightarrow dw=\sqrt{2}dv}{\Rightarrow }\int \frac{1}{\sqrt{2}(\frac{w^2}{2}+\frac{1}{2})}dw=\frac{1}{\sqrt{2}}\int \frac{1}{w^2+1}dw=\frac{1}{\sqrt{2}}{\tan }^{-1}w$$$$=\frac{1}{\sqrt{2}}{\tan }^{-1}(\sqrt{2}v+1)$$$$=\frac{\sqrt{2}+2}{2}\int \frac{2v+\sqrt{2}}{v^2+\sqrt{2}v+1}dv+(\sqrt{2}-\frac{\sqrt{2}+2}{\sqrt{2}})\int \frac{1}{v^2+\sqrt{2}v+1}dv$$$$=\frac{(\sqrt{2}+2)\ln (v^2+\sqrt{2}v+1)}{2}+\sqrt{2}(\sqrt{2}-\frac{\sqrt{2}+2}{\sqrt{2}}){\tan }^{-1}(\sqrt{2}v+1)$$$$......=\mathcal{H}\mathrm{oly}...\mathrm{god}...\mathrm{what}\mathrm{a}\mathrm{terrible}\mathrm{integral}\mathrm{question}$$

Commented by Tinku Tara last updated on 20/Dec/24

$$I={\int }_{\pi /6}^{\pi /3}\frac{1}{1+\sqrt{tanx}}dx....\left(1\right)$$$$\mathrm{substitute}u=\frac{\pi }{2}-x$$$$du=-dx$$$$x=\frac{\pi }{6}\Rightarrow u=\frac{\pi }{3}$$$$x=\frac{\pi }{3}\Rightarrow u=\frac{\pi }{2}$$$$I={\int }_{\pi /3}^{\pi /6}-\frac{1}{1+\sqrt{cotu}}du$$$$={\int }_{\pi /6}^{\pi /3}\frac{1}{1+\sqrt{cotu}}du={\int }_{\pi /6}^{\pi /3}\frac{\sqrt{tanu}}{1+\sqrt{tanu}}du$$$$={\int }_{\pi /6}^{\pi /3}\frac{\sqrt{tanx}}{1+\sqrt{tanx}}dx...\left(2\right)$$$$add\left(1\right)and\left(2\right)$$$$2I={\int }_{\pi /6}^{\pi /3}1dx=\frac{\pi }{6}\Rightarrow I=\frac{\pi }{12}$$

Commented by MathematicalUser2357 last updated on 26/Dec/24

Thanks��Equation editor owner

Answered by Simurdiera last updated on 20/Dec/24

$$Q.6Area=\pi {\mathrm{a}}^2$$

Answered by Tinku Tara last updated on 20/Dec/24

$$Q4$$$$put{tan}^{-1}x=u$$

Answered by TonyCWX08 last updated on 21/Dec/24

$$Q7.$$$$Area=\pi ab$$

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