<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Use the substitution $x=m\tan t$ to find $\int{\frac{1}{\sqrt{{{m}^{2}}+{{x}^{2}}}}\mathrm{d}x}$, where $m$ is a positive constant and $0\le t<\frac{\pi }{2}$.</span></div><p style="text-align: right;">[4]</p></li><li><div class="list-item-content"><span class="list-item-text">Find $\int{\frac{x}{\sqrt{{{m}^{2}}+{{x}^{2}}}}\mathrm{d}x}$.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Hence, find $\int{\frac{x}{\sqrt{{{m}^{2}}+{{x}^{2}}}}{{\tan }^{-1}}\left( \frac{x}{m} \right)}\,\mathrm{d}x$.</span></div><p style="text-align: right;">[3]</p></li></ol>

Use the substitution $x=m\tan t$ to find $\displaystyle\displaystyle \int{\frac{1}{\sqrt{{{m}^{2}}+{{x}^{2}}}}\mathrm{d}x}$, where $m$ is a positive constant and $0\le t<\frac{\pi }{2}$.

Find $\displaystyle\displaystyle \int{\frac{x}{\sqrt{{{m}^{2}}+{{x}^{2}}}}\mathrm{d}x}$.

Hence, find $\displaystyle\displaystyle \int{\frac{x}{\sqrt{{{m}^{2}}+{{x}^{2}}}}{{\tan }^{-1}}\left( \frac{x}{m} \right)}\,\mathrm{d}x$.

2022 NJC Promo Q6

2022 NJC Promo Q6

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