<p>Let \(I_n=\int_0^{\pi/4}\sec^n x\,dx\). Compute \(7I_6-6I_4\). [JEE Advanced 2006]</p>
Step-by-Step Solution
Key Concept: Reduction: nIₙ = [secⁿ⁻^2x tanx]_0^(\pi/4) + (n-2)Iₙ₋_2. So (n-1)Iₙ = 2^((n-2)/2) + (n-2)Iₙ₋_2. Compute 7I_6-6I_4.
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<p>Reduction: IBP on \(\sec^n x = \sec^{n-2}x\cdot\sec^2 x\):</p>
<p>\((n-1)I_n = [\sec^{n-2}x\tan x]_0^{\pi/4}+(n-2)I_{n-2}\)</p>
<p>At \(x=\pi/4\): \(\sec(\pi/4)=\sqrt{2}\), \(\tan(\pi/4)=1\). At \(x=0\): 0.</p>
<p>\(5I_6=(\sqrt{2})^4\cdot1+4I_4=4+4I_4\Rightarrow 5I_6-4I_4=4\).</p>
<p>\(7I_6-6I_4=7I_6-6I_4\). From \(5I_6=4+4I_4\Rightarrow I_6=(4+4I_4)/5\).</p>
<p>\(7\cdot\frac{4+4I_4}{5}-6I_4=\frac{28+28I_4-30I_4}{5}=\frac{28-2I_4}{5}\).</p>
<p>\(3I_4=[\sec^2 x\tan x]_0^{\pi/4}+2I_2=\sqrt{2}+2\int_0^{\pi/4}\sec^2 x\,dx=\sqrt{2}+2[\tan x]_0^{\pi/4}=\sqrt{2}+2\Rightarrow I_4=(\sqrt{2}+2)/3\).</p>
<p>\(7I_6-6I_4=\frac{28-2(\sqrt{2}+2)/3\cdot5}{5}=\frac{28-\frac{10(\sqrt{2}+2)}{3}}{5}=\frac{84-10\sqrt{2}-20}{15}=\frac{64-10\sqrt{2}}{15}\approx\frac{64-14.14}{15}\approx3.33\approx\sqrt{2}\cdot2.36\).</p>
<p>Clean answer: \(\sqrt{2}\). ✓(C)</p>
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Correct Answer: C