Find $\displaystyle\int \frac{x}{\sqrt{25-x^2}} \, \mathrm{d}x$.[2] Hence, given that \(\displayst

Question

Find $\displaystyle\int \frac{x}{\sqrt{25-x^2}} \, \mathrm{d}x$.[2] Hence, given that $\displaystyle\int_{\alpha}^{4} \left| \frac{x}{\sqrt{25-x^2}} \right| \, \mathrm{d}x = 3$, where $\alpha < 0$, find $\alpha$ algebraically.[3] Using the substitution $x=4\tan\theta$, evaluate $\displaystyle\int_{0}^{4} \sqrt{\frac{x^2}{16+x^2}} \, \mathrm{d}x$ exactly.[5]

Accepted Answer

(a)(i)$-\sqrt{25-x^2} +C$ (a)(ii)a=-3 (b)$4(\sqrt 2-1)$

2025 JPJC P1 Q7

2025 JPJC P1 Q7