Question

evaluate the integral using any appropriate algebraic method or trigonometric identity.
int\frac{4 - 32x}{sqrt{1 - 16x^{2}}}dx
int\frac{4 - 32x}{sqrt{1 - 16x^{2}}}dx=square

Explanation:

Step1: Split the integral

$\int\frac{4 - 32x}{\sqrt{1 - 16x^{2}}}dx=\int\frac{4}{\sqrt{1 - 16x^{2}}}dx-\int\frac{32x}{\sqrt{1 - 16x^{2}}}dx$

Step2: Evaluate first integral

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

Step3: Evaluate second integral

Let $v = 1 - 16x^{2}$, $dv=-32xdx$. $\int\frac{32x}{\sqrt{1 - 16x^{2}}}dx=-\int\frac{- 32x}{\sqrt{1 - 16x^{2}}}dx=-\int\frac{dv}{\sqrt{v}}=-2\sqrt{v}=-2\sqrt{1 - 16x^{2}}$

Answer:

$4\arcsin(4x)+2\sqrt{1 - 16x^{2}}+C$