Use the substitution $x=m\tan t$ to find \(\displaystyle\displaystyle \displaystyle\int{\frac{1}{\

Question

Use the substitution $x=m\tan t$ to find $\displaystyle\displaystyle \displaystyle\int{\frac{1}{\sqrt{{{m}^{2}}+{{x}^{2}}}}\mathrm{d}x}$, where $m$ is a positive constant and $0\le t<\frac{\pi }{2}$.[4] Find $\displaystyle\displaystyle \displaystyle\int{\frac{x}{\sqrt{{{m}^{2}}+{{x}^{2}}}}\mathrm{d}x}$.[2] Hence, find $\displaystyle\displaystyle \displaystyle\int{\frac{x}{\sqrt{{{m}^{2}}+{{x}^{2}}}}{{\tan }^{-1}}\left( \frac{x}{m} \right)}\,\mathrm{d}x$.[3]

Accepted Answer

(i) $\ln \left( x+\sqrt{{{m}^{2}}+{{x}^{2}}} ight)+{{c}_{1}}$ (ii) $\sqrt{{{m}^{2}}+{{x}^{2}}}+c$ (iii) $\sqrt{{{m}^{2}}+{{x}^{2}}}{{	an }^{-1}}\left( \frac{x}{m} ight)-m\ln \left( x+\sqrt{{{m}^{2}}+{{x}^{2}}} ight)+{{c}_{1}}$

2022 NJC Promo Q6

2022 NJC Promo Q6

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