121616949-math.177 - contains a polynomial expression that...

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8.3 Trigonometric Substitutions 163 EXAMPLE 8.8 Evaluate Z p 1 - x 2 dx . Let x = sin u so dx = cos u du . Then Z p 1 - x 2 dx = Z p 1 - sin 2 u cos u du = Z cos 2 u cos u du. We would like to replace cos 2 u by cos u , but this is valid only if cos u is positive, since cos 2 u is positive. Consider again the substitution x = sin u . We could just as well think of this as u = arcsin x . If we do, then by the definition of the arcsine, - π/ 2 u π/ 2, so cos u 0. Then we continue: Z cos 2 u cos u du = Z cos 2 u du = Z 1 + cos 2 u 2 du = u 2 + sin 2 u 4 + C = arcsin x 2 + sin(2 arcsin x ) 4 + C. This is a perfectly good answer, though the term sin(2 arcsin x ) is a bit unpleasant. It is possible to simplify this. Using the identity sin 2 x = 2 sin x cos x , we can write sin 2 u = 2 sin u cos u = 2 sin(arcsin x ) p 1 - sin 2 u = 2 x q 1 - sin 2 (arcsin x ) = 2 x 1 - x 2 . Then the full antiderivative is arcsin x 2 + 2 x 1 - x 2 4 = arcsin x 2 + x 1 - x 2 2 + C. This type of substitution is usually indicated when the function you wish to integrate
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Unformatted text preview: contains a polynomial expression that might allow you to use the fundamental identity sin 2 x + cos 2 x = 1 in one of three forms: cos 2 x = 1-sin 2 x sec 2 x = 1 + tan 2 x tan 2 x = sec 2 x-1 . If your function contains 1-x 2 , as in the example above, try x = sin u ; if it contains 1+ x 2 try x = tan u ; and if it contains x 2-1, try x = sec u . Sometimes you will need to try something a bit different to handle constants other than one. EXAMPLE 8.9 Evaluate Z p 4-9 x 2 dx . We start by rewriting this so that it looks more like the previous example: Z p 4-9 x 2 dx = Z p 4(1-(3 x/ 2) 2 ) dx = Z 2 p 1-(3 x/ 2) 2 dx....
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