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sgadi

  • 5 years ago

can someone solve int(e^x/(cos(x)))

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  1. anonymous
    • 5 years ago
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    Are you familiar with integration by parts?

  2. sgadi
    • 5 years ago
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    yes ... but i am not able to solve using that ...

  3. anonymous
    • 5 years ago
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    \[\int\limits_{?}^{?} udv = uv - \int\limits_{?}^{?} vdu\] What should we choose as our u and dv?

  4. sgadi
    • 5 years ago
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    may be v=e^x and u=cos(x)

  5. anonymous
    • 5 years ago
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    Sure. Carry out integration by parts once. What do you end up with? (You should have an integral remaining on the right hand side that seems just as difficult to solve as this one. Leave it be for now)

  6. anonymous
    • 5 years ago
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    Still there? You should end up with \[\int\limits\limits_{?}^{?}e ^{x}\cos x = e ^{x}\sin x - \int\limits\limits_{?}^{?}e ^{x}\sin x\] You would then perform integration by parts on the last term again, which will give you: \[\int\limits\limits_{?}^{?}e ^{x}\cos x = e ^{x}\sin x - (-e ^{x}\cos x + \int\limits_{?}^{?}e ^{x}\cos x )\] Distribute the negateive, and move the integral term to the right hand side. Divide both sides by 2.

  7. anonymous
    • 5 years ago
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    I thought he was trying to integrate \[\int\limits_{}e^x/cosx dx\]

  8. anonymous
    • 5 years ago
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    ...Guess that'll teach me to read the problem carefully. One moment to do the correct work.

  9. sgadi
    • 5 years ago
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    the problem is \[\int{\frac{e^x}{\cos{x}}}\]

  10. sgadi
    • 5 years ago
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    The actual problem I have is little more complex \[\int{\frac{e^{ax}}{\sqrt{b+c\cos{dx}}}}\]

  11. anonymous
    • 5 years ago
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    is the dx in the square root?

  12. sgadi
    • 5 years ago
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    I am sorry ... \[\int{\frac{e^{ax}}{\sqrt{b+c\cos{(x)}}}dx}\]

  13. anonymous
    • 5 years ago
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    What level of math is this for? I'm having difficulty finding a way to make this clean...

  14. sgadi
    • 5 years ago
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    I am working here to solve an model and got struck with this equation. It should be level of some master's engineering course.

  15. anonymous
    • 5 years ago
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    Well, I went ahead and used some software to examine the two deceptively simple looking integrals. The answer to the first involves a hypergeometric function, and it couldn't solve the second. Sorry, I don't think I'll be of much use to you on this.

  16. anonymous
    • 5 years ago
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    Well.... thats explains why I'm stummped

  17. anonymous
    • 5 years ago
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    \[\int\limits_{a}^{b}e^x/\cos x\] let u=e^x, du=e^x dx, dx=du/e^x so you have \[\int\limits_{a}^{b}du/cosx =\int\limits_{a}^{b} \sec x = \int\limits_{a}^{b}\ln \sec x + \tan x + C\] Hope this helps

  18. anonymous
    • 5 years ago
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    you cant do that because you have du/x, which means you have the change in u=du over cos(x), it has to either has to be du/u or dx/x or in the case of trig substitution dtheta/theta....... because of that your final answer lacks e^x and doesn't make sense

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