Use the supplied derivative, then split the sign · source reference

Printed question

(a) Write down $\frac d{dx}(\tan(x^2))$. Hence find $\int x^3\sec^2(x^2)dx$. [4]

(b) Evaluate $\int_0^4x^2|x-3|dx$ without the use of the graphic calculator. [3]

[2010/DHS/Prelim/I/6 (part)]

Printed working

(a)

$$\frac d{dx}\tan(x^2)=2x\sec^2(x^2).$$ $$\int x^3\sec^2(x^2)dx=\frac12\int x^2[2x\sec^2(x^2)]dx=\frac12\left[x^2\tan(x^2)-\int2x\tan(x^2)dx\right]=\frac12[x^2\tan(x^2)-\ln|\sec(x^2)|]+c.$$

(b)

$$|x-3|=\begin{cases}x-3,&x\ge3,\\-(x-3),&x<3.\end{cases}$$ $$\begin{aligned}\int_0^4x^2|x-3|dx&=-\int_0^3x^2(x-3)dx+\int_3^4x^2(x-3)dx\\&=-\int_0^3(x^3-3x^2)dx+\int_3^4(x^3-3x^2)dx\\&=-[x^4/4-x^3]_0^3+[x^4/4-x^3]_3^4=\frac{27}2.\end{aligned}$$

Teacher notes

Use integration by parts $\int u\,dv/dx\,dx=uv-\int v\,du/dx\,dx$, letting $u=x^2$, $dv/dx=2x\sec^2(x^2)$. The blue factor $2x\sec^2(x^2)$ matches the derivative supplied by the first task.

Printed answer key

(a) $\frac12[x^2\tan(x^2)-\ln|\sec(x^2)|]+c$; (b) $27/2$.