Q.10 Evaluate: ∫ x⁹·sec²(x¹⁰) dx

Answer: 1/10tan(x¹⁰) + c Put x¹⁰ = t , so 10x⁹ dx = dt , i.e. x⁹ dx = dt/10 . I = 1/10∫ sec² t dt = 1/10tan t + c = 1/10tan(x¹⁰) + c .

Evaluate: \(\int x^9\cdot\sec^2(x^{10})\,dx\)

Q.10 Evaluate: \(\int x^9\cdot\sec^2(x^{10})\,dx\)

Doubtly Asked 28 Oct 2026 Board exam · March 2025 hsc 12th Save WhatsApp Report

Best answer

Answer: \(\dfrac{1}{10}\tan(x^{10}) + c\)

Put \(x^{10} = t\), so \(10x^9\,dx = dt\), i.e. \(x^9\,dx = \dfrac{dt}{10}\).

\(I = \dfrac{1}{10}\int \sec^2 t\,dt = \dfrac{1}{10}\tan t + c = \dfrac{1}{10}\tan(x^{10}) + c\).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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