Question

evaluate the integral using any appropriate algebraic method or trigonometric identity. (int\frac{7 - 98x}{sqrt{4 - 49x^{2}}}dx) (int\frac{7 - 98x}{sqrt{4 - 49x^{2}}}dx=square)

Explanation:

Step1: Split the integral

$\int\frac{7 - 98x}{\sqrt{4 - 49x^{2}}}dx=\int\frac{7}{\sqrt{4 - 49x^{2}}}dx-\int\frac{98x}{\sqrt{4 - 49x^{2}}}dx$

Step2: Solve first integral

Let $u = \frac{7x}{2}$, $du=\frac{7}{2}dx$. $\int\frac{7}{\sqrt{4 - 49x^{2}}}dx = \int\frac{7}{\sqrt{4(1 - (\frac{7x}{2})^{2})}}dx=\int\frac{7}{2\sqrt{1 - u^{2}}}\cdot\frac{2}{7}du=\int\frac{1}{\sqrt{1 - u^{2}}}du=\arcsin(u)+C_1=\arcsin(\frac{7x}{2})+C_1$

Step3: Solve second integral

Let $v = 4 - 49x^{2}$, $dv=-98xdx$. $\int\frac{98x}{\sqrt{4 - 49x^{2}}}dx=-\int\frac{dv}{\sqrt{v}}=-2\sqrt{v}+C_2=-2\sqrt{4 - 49x^{2}}+C_2$

Step4: Combine results

$\int\frac{7 - 98x}{\sqrt{4 - 49x^{2}}}dx=\arcsin(\frac{7x}{2}) + 2\sqrt{4 - 49x^{2}}+C$

Answer:

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