Porisma

Worked solution

Reverse reciprocal-substitution bounds and reuse an integral

(a)(i) For a>1a>1, use u=1/xu=1/x to show ∫1adx/(1+x2)=∫1/a1du/(1+u2)\int_1^a dx/(1+x^2)=\int_{1/a}^1du/(1+u^2). marks

(ii) Hence show tan⁡−1a+tan⁡−1(1/a)=k\tan^{-1}a+\tan^{-1}(1/a)=k, finding kk. marks

(b)(i) Show ∫01x3/(1+x2) dx=1/2−(1/2)ln⁡2\int_0^1x^3/(1+x^2)\,dx=1/2-(1/2)\ln2. marks

(ii) Without using a calculator, evaluate ∫01x2tan⁡−1x dx\int_0^1x^2\tan^{-1}x\,dx. marks

[2013/JJC/I/12]

CJC · Review problems · Question 7

Answer

k=π/2k=\pi/2; ∫01x2tan⁡−1x dx=π/12−1/6+ln⁡2/6\int_0^1x^2\tan^{-1}x\,dx=\pi/12-1/6+\ln2/6.

(a)(i) Transform both the integrand and the bounds

u=1/xu=1/x gives dx=−du/u2dx=-du/u^2, with x=1↦u=1x=1\mapsto u=1 and x=a↦u=1/ax=a\mapsto u=1/a. Thus ∫1adx/(1+x2)=−∫11/adu/(1+u2)=∫1/a1du/(1+u2)\int_1^a dx/(1+x^2)=-\int_1^{1/a}du/(1+u^2)=\int_{1/a}^1du/(1+u^2).

How did this part go?

(a)(ii) Evaluate the two arctangent differences

tan⁡−1a−π/4=π/4−tan⁡−1(1/a)\tan^{-1}a-\pi/4=\pi/4-\tan^{-1}(1/a), so tan⁡−1a+tan⁡−1(1/a)=π/2\tan^{-1}a+\tan^{-1}(1/a)=\pi/2 and k=π/2k=\pi/2.

How did this part go?

(b)(i) Divide before integrating

x3/(1+x2)=x−x/(1+x2)x^3/(1+x^2)=x-x/(1+x^2). Its integral from 00 to 11 is [x2/2−12ln⁡(1+x2)]01=1/2−(1/2)ln⁡2[x^2/2-\tfrac12\ln(1+x^2)]_0^1=1/2-(1/2)\ln2.

How did this part go?

(b)(ii) Integrate by parts and reuse the result

With u=tan⁡−1xu=\tan^{-1}x, dv=x2dxdv=x^2dx, the integral is [x3tan⁡−1x/3]01−13∫01x3/(1+x2) dx=π/12−1/6+(1/6)ln⁡2[x^3\tan^{-1}x/3]_0^1-\tfrac13\int_0^1x^3/(1+x^2)\,dx=\pi/12-1/6+(1/6)\ln2.

How did this part go?

Common mistakes

Changing the limits and also discarding the substitution minus sign without checking orientation changes the integral.

Read the hint