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Reduction formula: relation between I_n and an earlier I for a definite integral

This type of question, asked 4 times: 14 May 2024 S2, 17 Apr 2026 S1, 20 Apr 2026 S2, 12 May 2026 S1. All 4 questions below, oldest first.

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  1. Question 1

    14 May 2024 S2

    142
    If In=∫0π/4tan⁡nθ dθI_n = \int_0^{\pi/4} \tan^n\theta\,d\theta, then I12+I10I_{12} + I_{10} =

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  2. Question 2

    17 Apr 2026 S1

    84
    If In=∫1e(log⁡ex)n dxI_n = \int_1^e(\log_e x)^n\,dx where n∈Z+n \in Z^+ and Im+mI2026=eI_m + mI_{2026} = e, then m=m =

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  3. Question 3

    20 Apr 2026 S2

    84
    If In=∫0π/4tan⁡nx dx,n∈NI_n = \int_0^{\pi/4}\tan^n x\,dx, n \in N then In+2+InI_{n+2} + I_n is equal to

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  4. Question 4

    12 May 2026 S1

    82
    If nn is an odd natural number and In=∫01ex(x−1)n dxI_n = \int_0^1 e^x(x - 1)^n\,dx, then the value of In+nIn−1I_n + nI_{n-1} is equal to ...

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