(CIE 0606/2021/m/22/Q11)
$\begin{array}{lll}\mathbf{11} & \text { (a) (i) Find } &\displaystyle \int\dfrac{1}{(10 x-1)^{6}} \mathrm{~d} x \text { . }\end{array}$
(ii) Find $\displaystyle\int \dfrac{\left(2 x^{3}+5\right)^{2}}{x} \mathrm{~d} x$.
(b) (i) Differentiate $y=\tan (3 x+1)$ with respect to $x$.
(ii) Hence find $\displaystyle\int_{\dfrac{\pi}{12}}^{\dfrac{\pi}{10}}\left(\dfrac{\sec ^{2}(3 x+1)}{2}-\sin x\right) \mathrm{d} x$.
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*********math solution************* $\begin{array}{ll}(a)(i) \displaystyle\int \dfrac{1}{(10 x-1)^{6}} d x &=\displaystyle\int(10 x-1)^{-6} d x \\&=\dfrac{1}{10} \times \dfrac{(10 x-1)^{-5}}{-5}+c \\&=-\dfrac{1}{50}(10 x-1)^{-5}+c \\\text { (ii) } \displaystyle\int \dfrac{\left(2 x^{3}+5\right)^{2}}{x} d x &=\displaystyle\int \dfrac{4 x^{6}+20 x^{3}+25}{x} d x \\&=\displaystyle\int\left(4 x^{5}+20 x^{2}+\dfrac{25}{x}\right) d x \\&=4 \dfrac{x^{6}}{6}+20 \dfrac{x^{3}}{3}+25 \ln x+c\end{array}$ b(i) $\dfrac{dy}{dx}=3\sec^2(3x+1)$ \text { (b) (ii) } $\begin{array}[t]{ll}\displaystyle\int_{\frac{\pi}{12}}^{\frac{\pi}{10}} &\left(\dfrac{\sec ^{2}(3 x+1)}{2}-\sin x\right) d x \\=& \dfrac{1}{2} \left[\dfrac{\tan (3 x+1)}{3}\right]_{\frac{\pi}{12}}^{\frac{\pi}{10}}+\left[\cos x\right]_{\frac{\pi}{12}}^{\frac{\pi}{10}} \\=& \dfrac{1}{6}\left[\tan \left(\dfrac{3 \pi}{10}+1\right)-\tan \left(\dfrac{3 \pi}{12}+1\right)\right] \\&+\left[\cos \left(\dfrac{\pi}{10}\right)-\cos \left(\dfrac{\pi}{12}\right)\right] \\=& 0.322\end{array}$ |
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