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nubeer

  • 4 years ago

integrate (1/(x^2+8x +25 )^0.5)

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  1. Andras
    • 4 years ago
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    This is a hard one x^2+8x+25=(x+4)^2+9 so use sub u=x+4 du=dx New integral: 1/(u^2+9)^0.5

  2. nubeer
    • 4 years ago
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    hmmm ya i also reached at this point but what after this?

  3. Andras
    • 4 years ago
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    thinking of the next substitution, maybe some trig function

  4. nubeer
    • 4 years ago
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    hmm yes i think that would be tan but how and why

  5. Andras
    • 4 years ago
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    I know! Recall that tan^2(x)+1=sec^2(x)

  6. Andras
    • 4 years ago
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    so we need to make 9tan^2(x) from u to get 9(tan^2(x)+1)

  7. Andras
    • 4 years ago
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    let u=3tan(z) du=3sec^2(z) dz

  8. Andras
    • 4 years ago
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    So the integral will be: 3sec^2(z)/3sec(z)=sec(z)

  9. Andras
    • 4 years ago
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    What the integral of secz? log(tanz+secz) (just looked it up)

  10. Andras
    • 4 years ago
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    we have u=3tanz expressing z=tan^-1(u/3)

  11. nubeer
    • 4 years ago
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    thanks man.. well pretty much hit the answer but can u explain this 1 step let u=3tan(z) du=3sec^2(z) dz.. why we suppose 3 tan(z) why not just tan

  12. Andras
    • 4 years ago
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    because if it is 3tan than (3tan)^2=9tan^2(x) so we can factor out 9 to get 9(tan^2(x)+1)

  13. nubeer
    • 4 years ago
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    ohh so just that was the reason to put 3?

  14. Andras
    • 4 years ago
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    yes

  15. Andras
    • 4 years ago
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    Hope it helped, now you just need to plug back the z and u to get x

  16. FoolForMath
    • 4 years ago
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    do you know about hyperbolic functions ?

  17. FoolForMath
    • 4 years ago
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    sinhx ?

  18. FoolForMath
    • 4 years ago
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    and this is a elementary problem no need to think hard on this one

  19. Andras
    • 4 years ago
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    e^x+e^-x/2 how would that help?

  20. Zarkon
    • 4 years ago
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    \[\frac{d}{dx}\operatorname{arsinh}(x) =\frac{1}{\sqrt{x^{2}+1}}\]

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