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A hint to get started

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

The reciprocal substitution reverses the bounds; use the earlier rational integral in the final integration by parts.

Read the worked solution