p218 ex22a q25

$\displaystyle \left[\frac{1}{16}sinx\right]_{0}^{tan^{-1}\frac{3}{4}}= 0.0375 =

\frac{3}{80}$

i can only do it by using calculator. i want to know if i can do it manually

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- Nov 5th 2007, 12:57 AMafeasfaerw23231233definite integrals question
p218 ex22a q25

$\displaystyle \left[\frac{1}{16}sinx\right]_{0}^{tan^{-1}\frac{3}{4}}= 0.0375 =

\frac{3}{80}$

i can only do it by using calculator. i want to know if i can do it manually - Nov 5th 2007, 01:26 AMticbol
= (1/16)[sin(arctan(3/4) -sin(arctan(0)]

arctan(3/4).

opp side = 3

adj side = 4

So,

hypotenuse = 5

And so,

sin(arctan(3/4)) = opp/hyp = 3/5

arctan(0).

opp side = 0

adj side = 1

So,

hypotenuse = 1 -----actually, no hypotenuse.

And so,

sin(arctan(0)) = opp/"hyp" = 0/1 = 0 -----this is only for comparison.

Of course, if tan(tetha) = 0, then theta = 0

And sin(0) = 0 also.

Hence,

= (1/16)[3/5 -0]

= 3/80