Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

How do I integrate this?

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  • alex
    Lv 7
    1 month ago
    Favorite Answer

    Hint:

    1 + ln(cosy)^(3/2) = 1+(3/2)ln(cosy)let u =  1+(3/2)ln(cosy)

  • Ian H
    Lv 7
    1 month ago

    Consider T = ln{ln[(cosy)^(3/2) + 1]}

    dT/dy = (-3/2)* tan y/{ln[(cosy)^(3/2) + 1]}

    so, I = ∫tan y/{ln[(cosy)^(3/2) + 1]}dy = -2/3T + C

    I = (-2/3)ln{ln[(cosy)^(3/2) + 1]} + C

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