Question

evaluate the integral using any appropriate algebraic method or trigonometric identity. (int\frac{4 - 32x}{sqrt{49 - 16x^{2}}}dx=)

Explanation:

Step1: Rewrite the denominator

We have $\sqrt{49 - 16x^{2}}=\sqrt{49(1-\frac{16}{49}x^{2})}=7\sqrt{1 - (\frac{4}{7}x)^{2}}$. Let $u=\frac{4}{7}x$, then $du=\frac{4}{7}dx$.

Step2: Rewrite the integral

The integral $\int\frac{4 - 32x}{\sqrt{49 - 16x^{2}}}dx=\int\frac{4}{\sqrt{49 - 16x^{2}}}dx-\int\frac{32x}{\sqrt{49 - 16x^{2}}}dx$. For $\int\frac{4}{\sqrt{49 - 16x^{2}}}dx$, substituting as above gives $\int\frac{4}{7\sqrt{1 - (\frac{4}{7}x)^{2}}}dx=\int\frac{1}{\sqrt{1 - u^{2}}}du=\arcsin(u)+C_1=\arcsin(\frac{4}{7}x)+C_1$. For $\int\frac{32x}{\sqrt{49 - 16x^{2}}}dx$, let $t = 49-16x^{2}$, then $dt=-32xdx$, and $\int\frac{32x}{\sqrt{49 - 16x^{2}}}dx=-\int\frac{dt}{\sqrt{t}}=-2\sqrt{t}+C_2=-2\sqrt{49 - 16x^{2}}+C_2$.

Step3: Combine the results

The original integral is $\arcsin(\frac{4}{7}x)+2\sqrt{49 - 16x^{2}}+C$.

Answer:

$\arcsin(\frac{4}{7}x)+2\sqrt{49 - 16x^{2}}+C$