Check whether the given equality can hold · source reference

Printed question

(i) Find $\frac d{dx}(\sin^{-1}x^2)$. [1]

(ii) Evaluate $\int_1^5\ln(\tan^{-1}(x/10))dx$ numerically. [1]

(iii) Given $0<k\le1$ and

$$\int_0^kx^2\sin^{-1}x^2dx=\int_1^5\ln(\tan^{-1}(x/10))dx,$$

show that $\int_0^k\frac{x^4}{\sqrt{1-x^4}}dx=ak^3\sin^{-1}k^2+b$, where constants $a,b$ are to be determined. [4]

[2010/RVHS/Prelim/I/4]

Printed working

(i)

$$\frac d{dx}(\sin^{-1}x^2)=\frac{2x}{\sqrt{1-x^4}}.$$

(ii)

$$\int_1^5\ln(\tan^{-1}(x/10))dx=-5.292774105=-5.29\quad\text{(3 s.f.)}.$$

(iii): printed conditional algebra

$$\int_0^kx^2\sin^{-1}x^2dx=[x^3\sin^{-1}x^2/3]_0^k-\int_0^k\frac23\frac{x^4}{\sqrt{1-x^4}}dx=\frac{k^3}3\sin^{-1}k^2-\int_0^k\frac23\frac{x^4}{\sqrt{1-x^4}}dx.$$ $$\frac{k^3}3\sin^{-1}k^2-\int_0^k\frac23\frac{x^4}{\sqrt{1-x^4}}dx=-5.292774105.$$ $$\int_0^k\frac{x^4}{\sqrt{1-x^4}}dx=\frac32\left[\frac{k^3}3\sin^{-1}k^2+5.292774105\right]=\frac12k^3\sin^{-1}k^2+7.94.$$

Printed constants: $a=0.5,b=7.94$.

Teacher notes

Not in MF27 (use chain rule): $\frac d{dx}\sin^{-1}f(x)=\frac{f'(x)}{\sqrt{1-(f(x))^2}}$.

From GC: 2nd → mode (blank screen), alpha → window → 4:fnInt(. Key in all the unknowns and press enter to obtain the answer.

Use integration by parts, $u=\sin^{-1}(x^2)$, $dv/dx=x^2$.

Printed answer key

(i) $2x/\sqrt{1-x^4}$; (ii) $-5.29$ (3 s.f.); (iii) $a=0.5,b=7.94$.

Mathematical review

The printed premise has no real solution $0<k\le1$: its left side is positive, while the right side is negative because $0<\tan^{-1}(x/10)<1$ for $1\le x\le5$. The displayed constants are formal consequences of the inconsistent premise, not an attainable value of $k$. The exact formal constant is $b=-(3/2)\int_1^5\ln(\tan^{-1}(x/10))dx$, approximately $7.9391611575$.