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    I have been set a question on integration, but I am having difficulty finding the correct solution.
    If I put the equation into my fx9860 or wolfram alpha I get a solution of 4.345 radians.
    However when I do the integration manually I come up with 2.367 radians.
    Here is my working out :-
    \displaystyle \int_0^\frac{\pi}{4} \left(5sin3x+3cos2x\right) dx = \left[ \frac{3}{2}sin2x-\frac{5}{3}cos3x \right]_0^\frac{\pi}{4}
    \displaystyle \left( \frac{3}{2}sin2\left(\frac{\pi}{  4}\right)-\frac{5}{3}cos3\left(\frac{\pi}{  4}\right) \right)-\left(\frac{3}{2}sin2\left(0 \right)-\frac{5}{3}cos3 \left(0\right)\right)
    (1.071--1.296)-(0)
    2.367 rads

    Where am I going wrong?

    Barney
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    When you've integrate the -cos, the sin term is negative (not positive as you've done)
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    The integral of cos(ax) is sin(ax)/a so you're just missing out your negative sign before the 3sin2x/2


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    (Original post by Barney63)
    (1.071--1.296)-(0)
    2.367 rads

    Where am I going wrong?

    Barney
    Cos (0) = 1 not 0

    and the answer is a number not an angle
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    (Original post by TenOfThem)
    Cos (0) = 1 not 0

    and the answer is a number not an angle
    It asks for the answer in radians.
    When cos(0)=1 that still only puts the answer to 3.367 rads.

    There was a typo on the equation when I first posted it but it was corrected.
    I originally posted 5Sin3x-3Cos2x where it should have been 5Sin3x+3Cos2x, that is why there isn't a minus sign in front of 3Sin2x/2.

    Barney
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    (Original post by Barney63)
    It asks for the answer in radians.
    When cos(0)=1 that still only puts the answer to 3.367 rads.

    There was a typo on the equation when I first posted it but it was corrected.
    I originally posted 5Sin3x-3Cos2x where it should have been 5Sin3x+3Cos2x, that is why there isn't a minus sign in front of 3Sin2x/2.

    Barney
    Check what you put into the calculator. I use the same values and get the correct answer.
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    (Original post by Barney63)
    I have been set a question on integration, but I am having difficulty finding the correct solution.
    If I put the equation into my fx9860 or wolfram alpha I get a solution of 4.345 radians.
    However when I do the integration manually I come up with 2.367 radians.
    Here is my working out :-
    \displaystyle \int_0^\frac{\pi}{4} \left(5sin3x+3cos2x\right) dx = \left[ \frac{3}{2}sin2x-\frac{5}{3}cos3x \right]_0^\frac{\pi}{4}
    \displaystyle \left( \frac{3}{2}sin2\left(\frac{\pi}{  4}\right)-\frac{5}{3}cos3\left(\frac{\pi}{  4}\right) \right)-\left(\frac{3}{2}sin2\left(0 \right)-\frac{5}{3}cos3 \left(0\right)\right)
    (1.071--1.296)-(0)
    2.367 rads

    Where am I going wrong?

    Barney
    3/2Sin(pi/2) = 3/2.
    5/3Cos(3pi/4) = -1.1785

    Thus 3/2Sin(pi/2) - 5/3Cos(3pi/4) = 2.6785

    3/2Sin(0) = 0
    5/3Cos(0) = 5/3.

    2.6785 -(-5/3)
    = 4.35 radians
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    \displaystyle \int_0^\frac{\pi}{4} \left(5sin3x+3cos2x\right) dx = \left[ \frac{3}{2}sin2x-\frac{5}{3}cos3x \right]_0^\frac{\pi}{4}

    = \left(\frac{3}{2}sin(\frac{\pi}{  2})-\frac{5}{3}cos(\frac{3\pi}{4}) \right) - \left(0 - \frac{5}{3} \right)

    = \left(\frac{3}{2} + \frac{5}{3\sqrt{2}} + \frac{5}{3} \right)

     = \frac{1}{6}\left(19 + 5 \sqrt{2} \right)
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    (Original post by Barney63)
    It asks for the answer in radians.
    Where does this question come from? You use radians to evaluate trig functions in calculus, but as TenOfThem pointed out, the answer to the integral is just a number - it has no "units of measurement".
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    (Original post by davros)
    Where does this question come from? You use radians to evaluate trig functions in calculus, but as TenOfThem pointed out, the answer to the integral is just a number - it has no "units of measurement".
    It is homework from college.

    Barney
 
 
 
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