antiderivative of 7^cost sint dt

**genlovesmusic09** · September 12th 2009, 01:47 PM

∫(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

**chengbin** · September 12th 2009, 01:55 PM

Have you tried rewriting the original function as

$\int \sqrt 7^{\sin 2t}dt$ ?

**Krizalid** · September 12th 2009, 02:03 PM

i'm pretty sure the integrand is actually $7^{\cos t}\sin t.$

**Plato** · September 12th 2009, 02:05 PM

Originally Posted by genlovesmusic09

∫(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

$\left. {\frac{{ - 7^{\cos (t)} }} {{\ln (7)}}} \right|_{t = 0}^{\frac{\pi } {2}} = \frac{{ - 7^0 }} {{\ln (7)}} - \frac{{ - 7^1 }} {{\ln (7)}} = \frac{6} {{\ln (7)}}$

**genlovesmusic09** · September 12th 2009, 02:31 PM

thanks! i excluded the 0 because i thought it would cancel out

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