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Qual b) \int \frac{x^{2}}{\sqrt{3-x^{2}}} d x\begin{aligned}\int \frac{\sqrt{3} \sin \theta)^{2}}{\sqrt{3-3 \sin ^{2} \theta}} \sqrt{3} \cos \theta d \theta & =\int \frac{3 \sin ^{2} \theta \cdot \sqrt{2} \cos \theta d \theta}{\sqrt{x} \cos \theta} \\& =\int 3 \sin ^{2} \theta d \theta \\& =\int 3\left[\frac{1-\cos 2 \theta}{2}\right] d \theta \quad[\operatorname{cosing} i \text { dentity } \\& =\frac{3}{2} \int(1-\cos 2 \theta) d \theta \quad\left(-\cos 2 \theta=2 \sin ^{2} \theta\right] \\& =\frac{3}{2}\left[\frac{\left.\theta-\sin ^{2} 2 \theta\right]+c}{2}\right]\end{aligned}So, \int \frac{x^{2}}{\sqrt{3-x^{2}}} d x=\frac{3}{2}\left[\frac{\theta-\sin ^{2} 2 \theta}{2}\right]+c where \theta=\sin ^{-1} xc)\begin{array}{l}\int \frac{\sqrt{16 x^{2}-9}}{x} d x=4 \int \frac{\sqrt{x^{2}-(3 y)^{2}}}{x} d x \\\text { Let } x=\frac{3}{4} \sec \theta \Rightarrow d x=\frac{3}{4} \sec \theta \tan \theta d \theta \\\theta=\sec ^{-14 x} \\=4 \int \frac{\sqrt{\left(\frac{3}{4} \sec \theta\right)^{2}\left(\frac{3}{4}\right)^{2}}}{\frac{3 \cdot}{4} \cdot \sec \theta} \frac{3}{4} \sec \tan \theta d \theta \\=y \mid\left(\frac{3}{4}\right) \sqrt{\sec ^{2} \theta-1} \cdot \tan \theta d \theta=3 \int \tan \theta \tan \theta d \theta \\\text { [using } \\=3 \int \tan ^{2} \theta d \theta \\1+\tan 2 \theta=\sec ^{2} \theta \\=3 \int\left(\sec ^{2} \theta-1\right) d \theta \\=3 \int \sec ^{2} \theta+3 / d \theta+c=3 \tan \theta-3 \theta+c \\=3 \tan \theta-3 \theta+c \text { where } \theta=\sec ^{-14} \frac{k}{3} \\\end{array} ...