tex] ...

Questions

September 2022 1 17 Report

Evaluate the following integrals :
[tex]\displaystyle \sf \int \sqrt{x} \: tan \bigg \{2 {tan}^{ - 1} \bigg( \dfrac{ \sqrt{ \sqrt{1 + \sqrt{x}} + 1} \: - \sqrt{ \sqrt{1 + \sqrt{x} } - 1 } }{ \sqrt{ \sqrt{1 + \sqrt{x}} + 1} \: + \sqrt{ \sqrt{1 + \sqrt{x} } - 1 }} \bigg) \bigg \} \:[/tex]

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EngrJohann

Verified answer

$\large \mathcal{SOLUTION:}$

$\footnotesize \begin{array}{l} I = \displaystyle \sf \int \sqrt{x} \: \tan \bigg \{2 \tan^{ - 1} \bigg( \dfrac{ \sqrt{ \sqrt{1 + \sqrt{x}} + 1} \: - \sqrt{ \sqrt{1 + \sqrt{x} } - 1 } }{ \sqrt{ \sqrt{1 + \sqrt{x}} + 1} \: + \sqrt{ \sqrt{1 + \sqrt{x} } - 1 }} \bigg) \bigg \} \: dx \\ \\ \\ \textsf{Let } \displaystyle \sf \tan \theta = \dfrac{ \sqrt{ \sqrt{1 + \sqrt{x}} + 1} \: - \sqrt{ \sqrt{1 + \sqrt{x} } - 1 } }{ \sqrt{ \sqrt{1 + \sqrt{x}} + 1} \: + \sqrt{ \sqrt{1 + \sqrt{x} } - 1 }} \\ \\ \\ \textsf{Simplifying } \\ \\ \\ \displaystyle \sf \tan \theta = \dfrac{ \sqrt{ \sqrt{1 + \sqrt{x}} + 1} \: - \sqrt{ \sqrt{1 + \sqrt{x} } - 1 } }{ \sqrt{ \sqrt{1 + \sqrt{x}} + 1} \: + \sqrt{ \sqrt{1 + \sqrt{x} } - 1 }} \cdot \dfrac{ \sqrt{ \sqrt{1 + \sqrt{x}} + 1} \: - \sqrt{ \sqrt{1 + \sqrt{x} } - 1 } }{ \sqrt{ \sqrt{1 + \sqrt{x}} + 1} \: - \sqrt{ \sqrt{1 + \sqrt{x} } - 1 }} \\ \\ \\ \displaystyle \sf \tan \theta = \dfrac{ \sqrt{1 + \sqrt{x}} + 1 \: - 2 \sqrt{1 +\sqrt{x} - 1} + \sqrt{1 + \sqrt{x} } - 1 }{ \sqrt{1 + \sqrt{x}} + 1 \: - \big(\sqrt{1 + \sqrt{x} } - 1 \big)} \\ \\ \\ \displaystyle \tan \theta = \sf \dfrac{2\sqrt{1 + \sqrt{x}} - 2\sqrt{\sqrt{x}} }{2} \\ \\ \\ \displaystyle \sf \tan \theta = \sqrt{1 + \sqrt{x}} - \sqrt{\sqrt{x}} \\ \\ \\ \textsf{By Double Angle Formula for Tangent,} \\ \\ \displaystyle\sf \tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^2 \theta} \\ \\ \displaystyle \sf \tan 2\theta = \dfrac{2\big(\sqrt{1 + \sqrt{x}} - \sqrt{\sqrt{x}}\big)}{1 - \big(\sqrt{1 + \sqrt{x}} - \sqrt{\sqrt{x}}\big)^2} \\ \\ \displaystyle \sf \tan 2\theta = \dfrac{2\big(\sqrt{1 + \sqrt{x}} - \sqrt{\sqrt{x}}\big)}{1 - \big(1 + \sqrt{x} - 2\sqrt{\sqrt{x}}\sqrt{1+\sqrt{x}} + \sqrt{x}\big)} \\ \\ \displaystyle \sf \tan 2\theta = \dfrac{2\big(\sqrt{1 + \sqrt{x}} - \sqrt{\sqrt{x}}\big)}{1 - 1 - \sqrt{x} + 2\sqrt{\sqrt{x}}\sqrt{1+\sqrt{x}} - \sqrt{x}} \\ \\ \displaystyle \sf \tan 2\theta = \dfrac{2\big(\sqrt{1 + \sqrt{x}} - \sqrt{\sqrt{x}}\big)}{2\sqrt{\sqrt{x}}\big(\sqrt{1+\sqrt{x}} - \sqrt{\sqrt{x}}\big)} \\ \\ \displaystyle \sf \tan 2\theta = \dfrac{1}{\sqrt{\sqrt{x}}} = x^{-\frac{1}{4}} \\ \\ \\ I = \displaystyle \sf \int x^{\frac{1}{2}} \cdot x^{-\frac{1}{4}} \: dx = \displaystyle \sf \int x^{\frac{1}{4}} \: dx = \displaystyle \sf \dfrac{x^{\frac{5}{4}}}{\frac{5}{4}} + C \\ \\ \therefore I = \boxed{\displaystyle \sf \dfrac{4}{5} x^{\frac{5}{4}} + C}\quad\textit{Answer} \end{array}$