Question

evaluate the integral using any appropriate algebraic method or trigonometric identity.
int\frac{3 - 18x}{sqrt{16 - 9x^{2}}}dx
int\frac{3 - 18x}{sqrt{16 - 9x^{2}}}dx=square

Explanation:

Step1: Split the integral

$\int\frac{3 - 18x}{\sqrt{16 - 9x^{2}}}dx=\int\frac{3}{\sqrt{16 - 9x^{2}}}dx-\int\frac{18x}{\sqrt{16 - 9x^{2}}}dx$

Step2: Solve first integral

Let $u = \frac{3}{4}x$, $dx=\frac{4}{3}du$. $\int\frac{3}{\sqrt{16 - 9x^{2}}}dx = 3\int\frac{1}{\sqrt{16(1 - (\frac{3}{4}x)^{2})}}dx=\arcsin(\frac{3}{4}x)$

Step3: Solve second integral

Let $t = 16 - 9x^{2}$, $dt=-18xdx$. $\int\frac{18x}{\sqrt{16 - 9x^{2}}}dx=-2\sqrt{16 - 9x^{2}}$

Answer:

$\arcsin(\frac{3}{4}x)+2\sqrt{16 - 9x^{2}}+C$