integrate the following function.\n int_{0}^{0.4} \frac{4mathrm{d}x}{sqrt{4 - 7x^{2}}}

Explanation:

Step1: Factor out the constant

We know that $\int_{a}^{b}kf(x)dx = k\int_{a}^{b}f(x)dx$. Here $k = 4$, so the integral $\int_{0}^{0.4}\frac{4dx}{\sqrt{4 - 7x^{2}}}=4\int_{0}^{0.4}\frac{dx}{\sqrt{4 - 7x^{2}}}$.

Step2: Rewrite the denominator

Rewrite $\sqrt{4 - 7x^{2}}$ as $\sqrt{4(1-\frac{7}{4}x^{2})}=2\sqrt{1 - (\frac{\sqrt{7}}{2}x)^{2}}$. Then the integral becomes $4\int_{0}^{0.4}\frac{dx}{2\sqrt{1 - (\frac{\sqrt{7}}{2}x)^{2}}}=2\int_{0}^{0.4}\frac{dx}{\sqrt{1 - (\frac{\sqrt{7}}{2}x)^{2}}}$.

Step3: Use substitution

Let $u=\frac{\sqrt{7}}{2}x$, then $du=\frac{\sqrt{7}}{2}dx$ and $dx=\frac{2}{\sqrt{7}}du$. When $x = 0$, $u = 0$; when $x=0.4$, $u=\frac{\sqrt{7}}{2}\times0.4 = 0.2\sqrt{7}$. The integral is $2\int_{0}^{0.2\sqrt{7}}\frac{1}{\sqrt{1 - u^{2}}}\times\frac{2}{\sqrt{7}}du=\frac{4}{\sqrt{7}}\int_{0}^{0.2\sqrt{7}}\frac{du}{\sqrt{1 - u^{2}}}$.

Step4: Recall the integral formula

We know that $\int\frac{du}{\sqrt{1 - u^{2}}}=\arcsin(u)+C$. So $\frac{4}{\sqrt{7}}\int_{0}^{0.2\sqrt{7}}\frac{du}{\sqrt{1 - u^{2}}}=\frac{4}{\sqrt{7}}[\arcsin(u)]_{0}^{0.2\sqrt{7}}$.

Step5: Evaluate the definite - integral

$\frac{4}{\sqrt{7}}(\arcsin(0.2\sqrt{7})-\arcsin(0))=\frac{4}{\sqrt{7}}\arcsin(0.2\sqrt{7})$.

Answer:

$\frac{4}{\sqrt{7}}\arcsin(0.2\sqrt{7})$