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    Solved, no more help needed


    i have to integrate sin2xcos3x, im not sure how to go about it, or if there are any trig identities that will help me solve it that ive forgot?

    any help appreciated + rep
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    Integration by parts?
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    (Original post by Lewk)
    i have to integrate sin2xcos3x, im not sure how to go about it, or if there are any trig identities that will help me solve it that ive forgot?

    any help appreciated + rep
    Expand \sin (A+B) +\sin (A-B). Can you use this to rewrite \sin 2x \cos 3x?
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    ? sin(ax) sin(bx) dx =

    let:
    sin(ax) = u ? (a) cos(ax) dx = du
    sin(bx) dx = dv ? (-1/b) cos(bx) = v

    thus, integrating by parts, you get:
    ? u dv = u v - ? v du ?

    ? sin(ax) sin(bx) dx = (-1/b) sin(ax) cos(bx) - ? (-1/b) cos(bx) (a) cos(ax) dx ?
    ? sin(ax) sin(bx) dx = (-1/b) sin(ax) cos(bx) + (a/b) ? cos(bx) cos(ax) dx ?
    a further by parts integration being needed, let:

    cos(ax) = u ? (- a) sin(ax) dx = du

    cos(bx) dx = dv ? (1/b) sin(bx) = v

    (note: when two subsequent by parts intergrations take place, it's crucial not to switch u and dv roles, as they were set in the first integration; thus, since sin(ax) was chosen as "u" yielding (a) cos(ax) dx = du, then cos(ax) must be taken as "u" in the subsequent integration)

    then integrating by parts again, you get:
    ? u dv = u v - ? v du ?

    ? sin(ax) sin(bx) dx = (-1/b) sin(ax) cos(bx) + (a/b) ? cos(bx) cos(ax) dx ?
    ? sin(ax) sin(bx) dx = (-1/b) sin(ax) cos(bx) +
    (a/b) [(1/b) cos(ax) sin(bx) - ? (1/b) sin(bx) (- a) sin(ax) dx] ?

    ? sin(ax) sin(bx) dx = (-1/b) sin(ax) cos(bx) +
    (a/b) [(1/b) cos(ax) sin(bx) + (a/b) ? sin(bx) sin(ax) dx] ?

    ? sin(ax) sin(bx) dx = (-1/b) sin(ax) cos(bx) + (a/b²) cos(ax) sin(bx) +
    (a²/b²) ? sin(bx) sin(ax) dx ?
    thus, having got the same unknown integral at both sides, let's collect it at left side
    (as if it were a common algebraic unknown):

    ? sin(ax) sin(bx) dx - (a²/b²) ? sin(ax) sin(bx) dx = (-1/b) sin(ax) cos(bx) +
    (a/b²) cos(ax) sin(bx) + C ?

    [1 - (a²/b²)] ? sin(ax) sin(bx) dx = (-1/b) sin(ax) cos(bx) + (a/b²) cos(ax) sin(bx) + C ?
    [(b² - a²)/b²)] ? sin(ax) sin(bx) dx = (-1/b) sin(ax) cos(bx) + (a/b²) cos(ax) sin(bx) + C ?
    and therefore:

    ? sin(ax) sin(bx) dx = [b²/(b²- a²)] [(-1/b) sin(ax) cos(bx) + (a/b²) cos(ax) sin(bx)] + C ?
    and, finally, expanding,

    ? sin(ax) sin(bx) dx = [-b/(b²- a²)] sin(ax) cos(bx) + [a/(b²- a²)] cos(ax) sin(bx)] + C =

    [b/(a² - b²)] sin(ax) cos(bx) + [a/(b² - a²)] cos(ax) sin(bx)] + C

    I hope it helps...
    Bye!

    Fyi question marks are meant to be integral signs, and this is the longer method the substitution is much easier :£
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    sinAcosB\equiv\frac{1}{2}(sin(A+  B)+sin(A-B))
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    (Original post by Kevlar)
    .
    what the what
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    (Original post by Pheylan)
    sinAcosB\equiv\frac{1}{2}(sin(A+  B)+sin(A-B))
    :five:
    Thank God that you didn't suggest IBP...
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    (Original post by Pheylan)
    what the what
    derp derp its right yO! IBP=sexy <3
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    (Original post by Kevlar)
    ? sin(ax) sin(bx) dx =

    let:
    sin(ax) = u ? (a) cos(ax) dx = du
    sin(bx) dx = dv ? (-1/b) cos(bx) = v

    thus, integrating by parts, you get:
    ? u dv = u v - ? v du ?

    ? sin(ax) sin(bx) dx = (-1/b) sin(ax) cos(bx) - ? (-1/b) cos(bx) (a) cos(ax) dx ?
    ? sin(ax) sin(bx) dx = (-1/b) sin(ax) cos(bx) + (a/b) ? cos(bx) cos(ax) dx ?
    a further by parts integration being needed, let:

    cos(ax) = u ? (- a) sin(ax) dx = du

    cos(bx) dx = dv ? (1/b) sin(bx) = v

    (note: when two subsequent by parts intergrations take place, it's crucial not to switch u and dv roles, as they were set in the first integration; thus, since sin(ax) was chosen as "u" yielding (a) cos(ax) dx = du, then cos(ax) must be taken as "u" in the subsequent integration)

    then integrating by parts again, you get:
    ? u dv = u v - ? v du ?

    ? sin(ax) sin(bx) dx = (-1/b) sin(ax) cos(bx) + (a/b) ? cos(bx) cos(ax) dx ?
    ? sin(ax) sin(bx) dx = (-1/b) sin(ax) cos(bx) +
    (a/b) [(1/b) cos(ax) sin(bx) - ? (1/b) sin(bx) (- a) sin(ax) dx] ?

    ? sin(ax) sin(bx) dx = (-1/b) sin(ax) cos(bx) +
    (a/b) [(1/b) cos(ax) sin(bx) + (a/b) ? sin(bx) sin(ax) dx] ?

    ? sin(ax) sin(bx) dx = (-1/b) sin(ax) cos(bx) + (a/b²) cos(ax) sin(bx) +
    (a²/b²) ? sin(bx) sin(ax) dx ?
    thus, having got the same unknown integral at both sides, let's collect it at left side
    (as if it were a common algebraic unknown):

    ? sin(ax) sin(bx) dx - (a²/b²) ? sin(ax) sin(bx) dx = (-1/b) sin(ax) cos(bx) +
    (a/b²) cos(ax) sin(bx) + C ?

    [1 - (a²/b²)] ? sin(ax) sin(bx) dx = (-1/b) sin(ax) cos(bx) + (a/b²) cos(ax) sin(bx) + C ?
    [(b² - a²)/b²)] ? sin(ax) sin(bx) dx = (-1/b) sin(ax) cos(bx) + (a/b²) cos(ax) sin(bx) + C ?
    and therefore:

    ? sin(ax) sin(bx) dx = [b²/(b²- a²)] [(-1/b) sin(ax) cos(bx) + (a/b²) cos(ax) sin(bx)] + C ?
    and, finally, expanding,

    ? sin(ax) sin(bx) dx = [-b/(b²- a²)] sin(ax) cos(bx) + [a/(b²- a²)] cos(ax) sin(bx)] + C =

    [b/(a² - b²)] sin(ax) cos(bx) + [a/(b² - a²)] cos(ax) sin(bx)] + C

    I hope it helps...
    Bye!

    Fyi question marks are meant to be integral signs, and this is the longer method the substitution is much easier :£
    and after all that, you answered the question of how to integrate sin(ax)sin(bx) when the question was of the form sin(ax)cos(bx)

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    (Original post by ziedj)
    and after all that, you answered the question of how to integrate sin(ax)sin(bx) when the question was of the form sin(ax)cos(bx)

    LOL my bad i r hungover and didnt read it right XDDDD
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    thanks to the people who suggested the product-to-sum formula , question is done & its correct
    & thanks to kevlar too for taking the time to write all that even though it was wrong... :x
 
 
 
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