Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Calculus AP...Integration Help?

What is the integral of:

100t²sin(√t)

Its pretty tough....I need some help on this one.

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  • 1 decade ago
    Best Answer

    ∫ 100 t² sin(√t) dt =

    pull the constant out:

    100 ∫ t² sin(√t) dt =

    let (√t) = y

    t = y²

    dt = 2y dy

    substituting, you have:

    100 ∫ t² sin(√t) dt = 100 ∫ (y²)² siny 2y dy =

    100 (2) ∫ y^4 y siny dy =

    200 ∫ y^5 siny dy =

    let:

    y^5 = u → 5y^4 dy = du

    siny dy = dv → - cosy = v

    integrate by parts, yielding:

    ∫ u dv = v u - ∫ v du

    200 ∫ y^5 siny dy = 200 [y^5 (- cosy) - ∫ (- cosy) 5y^4 dy] =

    200 [- y^5 cosy + 5 ∫ y^4 cosy dy] =

    - 200y^5 cosy + 1000 ∫ y^4 cosy dy =

    integrate the remaining integral by parts again, letting:

    y^4 = u → 4y³ dy = du

    cosy dy = dv → siny = v

    yielding:

    - 200y^5 cosy + 1000 ( ∫ y^4 cosy dy) =

    - 200y^5 cosy + 1000 (y^4 siny - ∫ siny 4y³ dy) =

    - 200y^5 cosy + 1000 (y^4 siny - 4 ∫ y³ siny dy) =

    - 200y^5 cosy + 1000y^4 siny - 4000 ∫ y³ siny dy =

    further integration by parts being needed, let:

    y³ = u → 3y² dy = du

    siny dy = dv → - cosy = v

    obtaining:

    - 200y^5 cosy + 1000y^4 siny - 4000 [∫ y³ siny dy] =

    - 200y^5 cosy + 1000y^4 siny - 4000 [y³ (- cosy) - ∫ (- cosy) 3y² dy] =

    - 200y^5 cosy + 1000y^4 siny - 4000 (- y³ cosy + 3 ∫ y² cosy dy) =

    - 200y^5 cosy + 1000y^4 siny + 4000y³ cosy - 12000 ∫ y² cosy dy =

    let:

    y² = u → 2y dy = du

    cosy dy = dv → siny = v

    integrating by parts, you get:

    - 200y^5 cosy + 1000y^4 siny + 4000y³ cosy - 12000 ( ∫ y² cosy dy) =

    - 200y^5 cosy + 1000y^4 siny + 4000y³ cosy - 12000 (y² siny - ∫ siny 2y dy) =

    - 200y^5 cosy + 1000y^4 siny + 4000y³ cosy - 12000 (y² siny - 2 ∫ y siny dy) =

    - 200y^5 cosy + 1000y^4 siny + 4000y³ cosy - 12000y² siny + 24000 ∫ y siny dy =

    let:

    y = u → dy = du

    siny dy = dv → - cosy = v

    integrating by parts, you get:

    - 200y^5 cosy + 1000y^4 siny + 4000y³ cosy - 12000y² siny + 24000 [∫ y siny dy] =

    - 200y^5 cosy + 1000y^4 siny + 4000y³ cosy - 12000y² siny + 24000 [y (- cosy) -

    ∫ (- cosy) dy] =

    - 200y^5 cosy + 1000y^4 siny + 4000y³ cosy - 12000y² siny + 24000 (- y cosy +

    ∫ cosy dy) =

    - 200y^5 cosy + 1000y^4 siny + 4000y³ cosy - 12000y² siny - 24000 y cosy +

    24000 ∫ cosy dy =

    - 200y^5 cosy + 1000y^4 siny + 4000y³ cosy - 12000y² siny - 24000 y cosy +

    24000 siny + C

    finally, substitute back (√t) for y, yielding:

    - 200(√t)^5 cos(√t) + 1000(√t)^4 sin(√t) + 4000(√t)³ cos(√t) - 12000(√t)² sin(√t) -

    24000(√t) cos(√t) + 24000 sin(√t) + C

    ending with:

    ∫ 100 t² sin(√t) dt = - 200t²(√t)cos(√t) + 1000t² sin(√t) + 4000t(√t)cos(√t) -

    12000t sin(√t) - 24000(√t)cos(√t) + 24000 sin(√t) + C

    I hope it helps..

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