FA07Ma1aSol6%20(1)

FA07Ma1aSol6%20(1) - Math 1a Fall Term 2007 SOLUTIONS TO...

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Unformatted text preview: Math 1a Fall Term 2007 SOLUTIONS TO HOMEWORK 6 Problem 1 (20 points) (Apostol, pg 217, # 26) (a) Show that Z π xf (sin x ) dx = π 2 Z π f (sin x ) dx. [Hint: u = π- x ] (b) Use part (a) to deduce the formula Z π x sin x 1 + cos 2 x dx = π Z 1 dx 1 + x 2 . Solution 1(a) Let I = Z π xf (sin x ) dx. Substitute u = π- x ; then x = π- u and dx =- du . As x increases from 0 to π , u decreases from π to 0, thus I = Z π ( π- u ) f (sin( π- u ))(- du ) =- Z π ( π- u ) f (sin( π- u )) du. Using sin( π- u ) = sin u and the fact that for an integrable function g we have R b a g ( x ) dx =- R a b g ( x ) dx , we get I = Z π ( π- u ) f (sin u ) du = Z π πf (sin u ) du- Z π uf (sin u ) du. Replacing u by x we see that the second integral is I again, thus I = π Z π f (sin x ) dx- I, that is I = π 2 Z π f (sin x ) dx. Solution 1(b) Let I = Z π x sin x 1 + cos 2 x dx. Putting f ( t ) = t/ (2- t 2 ), f (sin x ) = sin x 2- sin 2 x = sin x 1 + cos 2 x and the integral can be rewritten as I = Z π xf (sin x ) dx. Using part (a) we get I = π 2 Z π f (sin x ) dx. 1 2 SOLUTIONS TO HOMEWORK 6 Rewriting back the expression for f (sin x ) we get I = π 2 Z π sin x 1 + cos 2 x dx. Substitute u =- cos x ; then du = sin x dx . Further, as x increases from 0 to π we see that u increases from- cos(0) =- 1 to- cos( π ) = 1. So I = π 2 Z 1- 1 du 1 + u 2 . Notice that g ( u ) = du 1+ u 2 is an even function of u , in other words g (- u ) = g ( u ). Since for even functions R a- a g ( u ) du = 2 R a g ( u ) du (split into two pieces, substitute u by- t in one of them, change back t by u ), we get I = ( π 2 )2 Z 1 du 1 + u 2 = π Z 1 du 1 + u 2 . Replacing the variable u by the variable x , I = π Z 1 dx 1 + x 2 , which is what we set out to deduce....
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FA07Ma1aSol6%20(1) - Math 1a Fall Term 2007 SOLUTIONS TO...

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