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Integral of x times a function symmetric about the midpoint, using x -> a + b - x to remove x

This type of question, asked 7 times: 26 Jun 2022, 31 Jan 2024, 1 Feb 2024, 5 Apr 2024, 23 Jan 2025, 3 Apr 2025, 7 Apr 2025. All 7 questions below, oldest first.

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  1. Question 1

    26 Jun 2022

    88
    The value of the integral 48π4∫0π(3πx22−x3)sin⁡x1+cos⁡2xdx\frac{48}{\pi^{4}}\int_{0}^{\pi} \left( \frac{3\pi x^{2}}{2}-x^{3} \right)\frac{\sin x}{1 +\cos^{2}x}dx is equal to
  2. Question 2

    31 Jan 2024

    171
    ∣120π3∫0πx2sin⁡xcosxsin⁡4x+cos⁡4xdx∣\left| \frac{120}{\pi^{3}}\int_{0}^{\pi} \frac{x^{2}\sin xcosx}{\sin^{4}x + \cos^{4}x}dx \right| is equal to.
  3. Question 3

    1 Feb 2024

    62
    The value of the integral ∫0π4xdxsin⁡4(2x)+cos⁡4(2x)\int_{0}^{\frac{\pi}{4}} \frac{xdx}{\sin^{4}(2x) +\cos^{4}(2x)} equals:

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  4. Question 4

    5 Apr 2024

    79
    The value of ∫−ππ2y(1+sin⁡y)1+cos⁡2ydy\int_{- \pi}^{\pi} \frac{2y(1 + \sin y)}{1 +\cos^{2}y}dy is :

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  5. Question 5

    23 Jan 2025

    144
    If I=∫0π2sin⁡32xsin⁡32x+cos⁡32xdxI =\int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}}x}{\sin^{\frac{3}{2}}x +\cos^{\frac{3}{2}}x}dx, then ∫02Ixsin⁡xcos⁡xsin⁡4x+cos⁡4xdx\int_{0}^{2I} \frac{x\sin x\cos x}{\sin^{4}x +\cos^{4}x}dx equals:

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  6. Question 6

    3 Apr 2025

    134
    The integral ∫0π8xdx4cos⁡2x+sin⁡2x\int_{0}^{\pi} \frac{8xdx}{4\cos^{2}x+\sin^{2}x} is equal to.

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  7. Question 7

    7 Apr 2025

    60
    The integral ∫0π(x+3)sin⁡x1+3cos⁡2xdx\int_{0}^{\pi} \frac{(x+ 3)\sin x}{1 + 3\cos^{2}x}dx is equal to :

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