Integration techniques: TMJC 2024 Promo Q7

(a)

Find $\displaystyle\int\frac{1}{\sqrt{4-x^2}}\,\,\mathrm{d} x$. [1]

(a) $$\begin{aligned} \int\frac{1}{\sqrt{4-x^2}}\,\,\mathrm{d} x &=\int\frac{1}{\sqrt{2^2-x^2}}\,\,\mathrm{d} x\\ &=\sin^{-1}\left(\frac{x}{2}\right)+C \end{aligned}$$

(b)

Find $\displaystyle\int\cos^2x\,\,\mathrm{d} x$. [2]

(b) $$\begin{aligned} \int\cos^2x\,\,\mathrm{d} x&=\frac12\int(1+\cos2x)\,\,\mathrm{d} x\\ &=\frac12\left(x+\frac{\sin2x}{2}\right)+C \end{aligned}$$ Remember to apply Golden Rule!

(c)

Using integration by parts, find $\displaystyle\int\tan^{-1}x\,\,\mathrm{d} x$. [3]

(c) $$\begin{aligned} \int\tan^{-1}x\,\,\mathrm{d} x &=x\tan^{-1}x-\int x\left(\frac{1}{1+x^2}\right)\,\,\mathrm{d} x\\ &=x\tan^{-1}x-\frac12\int\frac{2x}{1+x^2}\,\,\mathrm{d} x\\ &=x\tan^{-1}x-\frac12\ln\lvert1+x^2\rvert+C\\ &=x\tan^{-1}x-\frac12\ln(1+x^2)+C,\quad\text{since }1+x^2>0 \end{aligned}$$ Keep, Differentiate $\tan^{-1}x$. Integrate “1”.

(d)

Use the substitution $u=\sqrt{x-2}$ to find the exact value of $\displaystyle\int_2^4\frac{x+1}{x\sqrt{x-2}}\,\,\mathrm{d} x$. [4]

(d) $$\begin{aligned} \int_2^4\frac{x+1}{x\sqrt{x-2}}\,\,\mathrm{d} x &=\int_0^{\sqrt2}\frac{2(u^2+2)+1}{(u^2+2)u}(2u)\,\,\mathrm{d} u\\ &=2\int_0^{\sqrt2}\frac{u^2+3}{u^2+2}\,\,\mathrm{d} u\\ &=2\int_0^{\sqrt2}\left(1+\frac{1}{u^2+(\sqrt2)^2}\right)\,\,\mathrm{d} u\\ &=2\left[u+\frac{1}{\sqrt2}\tan^{-1}\left(\frac{u}{\sqrt2}\right)\right]_0^{\sqrt2}\\ &=2\left[\sqrt2+\frac{1}{\sqrt2}\tan^{-1}\left(\frac{\sqrt2}{\sqrt2}\right)-0\right]\\ &=2\left[\sqrt2+\frac{1}{\sqrt2}\left(\frac{\pi}{4}\right)\right]\\ &=\sqrt2\left(2+\frac{\pi}{4}\right) \end{aligned}$$

$u=\sqrt{x-2}$

$u^2=x-2$

Differentiating with respect to $x$,$2u\dfrac{\,\mathrm{d} u}{\,\mathrm{d} x}=1$

When $x=4$, $u=\sqrt{4-2}=\sqrt2$

When $x=2$, $u=\sqrt{2-2}=0$

Remember to change the limits too!

Ensure that all substitutions are completed in 1 step to reduce careless mistakes.

It is recommended to express the integral in standard form first before applying MF27 integration formula. This usually reduces the chance of making careless mistakes.