Check whether the given equality can hold

(i) Find ddx(sin⁡−1x2)\frac d{dx}(\sin^{-1}x^2). mark

(ii) Evaluate ∫15ln⁡(tan⁡−1(x/10))dx\int_1^5\ln(\tan^{-1}(x/10))dx numerically. mark

(iii) Given 0<k≤10<k\le1 and

∫0kx2sin⁡−1x2dx=∫15ln⁡(tan⁡−1(x/10))dx,\int_0^kx^2\sin^{-1}x^2dx=\int_1^5\ln(\tan^{-1}(x/10))dx,

show that ∫0kx41−x4dx=ak3sin⁡−1k2+b\int_0^k\frac{x^4}{\sqrt{1-x^4}}dx=ak^3\sin^{-1}k^2+b, where constants a,ba,b are to be determined. marks

[2010/RVHS/Prelim/I/4]

Total marks: 6

Solution

(i)

Derivative of an inverse trig function

Find ddx(sin⁡−1x2)\dfrac d{dx}\left(\sin^{-1}x^2\right).

Hint

  1. Chain rule: ddxsin⁡−1u=u′1−u2\frac d{dx}\sin^{-1}u=\frac{u'}{\sqrt{1-u^2}} with u=x2u=x^2.

Answer · (i)

2x/1−x42x/\sqrt{1-x^4}.

−1<x<1-1<x<1.

(ii)

Numerical integration

Evaluate ∫15ln⁡(tan⁡−1x10)dx\int_1^5\ln\left(\tan^{-1}\frac x{10}\right)dx numerically.

Explore this idea

Hint

  1. Use the calculator's definite-integral function in radian mode.

Answer · (ii)

J≈−5.29J\approx-5.29 (3 s.f.).

Explore the idea

A negative integrand

Drag x to explore the numerical signed integral over the selected interval.

Upper limit xx = 5.00
Try an example

How the shaded areas add up

The limits run left to right. Area above the axis adds; area below the axis subtracts.

Above-axis contribution
0
Below-axis contribution
-5.293
Signed integral
-5.293

All three bars share one scale. Positive contributions extend right of zero; negative contributions extend left.

Signed integral from 1.00 to x
-5.293
Compare the running integral on a graph
  • Integrand
  • Numerical signed integral from the start to x
  • Part of the integrand above the axis
  • Part of the integrand below the axis
Key idea

Every value of the curve is below the axis, so the area (the integral) is negative: −5.29-5.29.

Ideas to try
The integrand is negative everywhere on [1,5][1,5]: what sign must the integral have?
(iii)

Checking feasibility

Given 0<k≤10<k\le1 and ∫0kx2sin⁡−1x2 dx=∫15ln⁡(tan⁡−1x10)dx\int_0^kx^2\sin^{-1}x^2\,dx=\int_1^5\ln\left(\tan^{-1}\frac x{10}\right)dx, show that ∫0kx41−x4 dx=ak3sin⁡−1k2+b\int_0^k\dfrac{x^4}{\sqrt{1-x^4}}\,dx=ak^3\sin^{-1}k^2+b and find aa and bb.

Explore this idea

Hint

  1. Integrate ∫0kx2sin⁡−1x2 dx\int_0^kx^2\sin^{-1}x^2\,dx by parts using part (i).

  2. Equate with −5.29-5.29 and rearrange.

  3. Then ask: can the premise hold at all?

Answer · (iii)

No admissible kk satisfies the stated premise.

The left integral is positive but J<0. Formal rearrangement gives a=1/2,b=−3J/2≈7.94, but these are only a conditional identity with an impossible premise.

Explore the idea

The left-hand integral is positive

Drag x to explore the numerical signed integral over the selected interval.

Upper limit xx = 1.00
Try an example

How the shaded areas add up

The limits run left to right. Area above the axis adds; area below the axis subtracts.

Above-axis contribution
0.232
Below-axis contribution
0
Signed integral
0.232

All three bars share one scale. Positive contributions extend right of zero; negative contributions extend left.

Signed integral from 0.00 to x
0.232
Compare the running integral on a graph
  • Integrand
  • Numerical signed integral from the start to x
  • Part of the integrand above the axis
  • Part of the integrand below the axis
Key idea

For 0<k≤10<k\le1, the left integral is positive and at most its value at k=1k=1, about 0.232. The right integral is negative, about −5.29. The equality is impossible throughout the stated domain.

Ideas to try
Drag xx to 1. Could this area ever equal −5.29-5.29?