Integrate the following functions with respect to $x$:

(i) $\dfrac{1}{\sqrt{1-(4x)^{2}}}$

(ii) $\dfrac{1}{\sqrt{1-81x^{2}}}$

(iii) $\dfrac{1}{1+36x^{2}}$

Question

Integrate the following functions with respect to $x$:

(i) $\dfrac{1}{\sqrt{1-(4x)^{2}}}$

(ii) $\dfrac{1}{\sqrt{1-81x^{2}}}$

(iii) $\dfrac{1}{1+36x^{2}}$

Accepted Answer

> [!TLDR]
>
> Recognise each denominator as $1-(ax)^2$ or $1+(ax)^2$ (pulling the coefficient inside the square where needed), then apply $\int\frac{dx}{\sqrt{1-(ax)^2}}=\frac1a\sin^{-1}(ax)+c$ or $\int\frac{dx}{1+(ax)^2}=\frac1a	an^{-1}(ax)+c$.

**Step 1. Part (i).** Already in the form $1-(ax)^2$ with $a=4$.
$$\int \frac{dx}{\sqrt{1-(4x)^2}}=\frac14\sin^{-1}(4x)+c.$$

**Step 2. Part (ii).** Recognise $81x^2=(9x)^2$, so $a=9$.
$$\int \frac{dx}{\sqrt{1-81x^2}}=\int\frac{dx}{\sqrt{1-(9x)^2}}=\frac19\sin^{-1}(9x)+c.$$

**Step 3. Part (iii).** Recognise $36x^2=(6x)^2$, so $a=6$.
$$\int \frac{dx}{1+36x^2}=\int\frac{dx}{1+(6x)^2}=\frac16	an^{-1}(6x)+c.$$ …

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Q.Integrate the following functions with respect to $x$ :

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