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Math Help - antiderivative of 7^cost sint dt

  1. #1
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    Post antiderivative of 7^cost sint dt

    (from 0 to pi/2) 7^cost sint dt
    i know that it is a^u du
    but sin needs to be negative so I tried this:
    -(from 0 to pi/2) 7^cost (-sint) dt
    then made u=cost and du= -sint
    so: 7^u du= 7^u/ ln7= 7^cost /ln7
    and when i plug in pi/2 i get 1/ln7
    but the answer is 6/ln7
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  2. #2
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    Have you tried rewriting the original function as

    \int \sqrt 7^{\sin 2t}dt?
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  3. #3
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    i'm pretty sure the integrand is actually 7^{\cos t}\sin t.
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  4. #4
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    Quote Originally Posted by genlovesmusic09 View Post
    (from 0 to pi/2) 7^cost sint dt
    i know that it is a^u du
    but sin needs to be negative so I tried this:
    -(from 0 to pi/2) 7^cost (-sint) dt
    then made u=cost and du= -sint
    so: 7^u du= 7^u/ ln7= 7^cost /ln7
    and when i plug in pi/2 i get 1/ln7
    but the answer is 6/ln7
    \left. {\frac{{ - 7^{\cos (t)} }}<br />
{{\ln (7)}}} \right|_{t = 0}^{\frac{\pi }<br />
{2}}  = \frac{{ - 7^0 }}<br />
{{\ln (7)}} - \frac{{ - 7^1 }}<br />
{{\ln (7)}} = \frac{6}<br />
{{\ln (7)}}
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  5. #5
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    thanks! i excluded the 0 because i thought it would cancel out
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