Reversing the Chain Rule. As we have seen from the Second Fundamental Theorem ( 4.3), the easiest way to evaluate an integral b


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1 Mth 32 Substitution Method Stewrt 4.5 Reversing the Chin Rule. As we hve seen from the Second Fundmentl Theorem ( 4.3), the esiest wy to evlute n integrl b f(x) dx is to find n ntiderivtive, the indefinite integrl f(x) dx F (x) + C, so tht b f(x) dx F (b) F (). Building on 3.9, we will lern severl methods to find ntiderivtives which reverse our methods of differentition, in this cse the Chin Rule. For exmple, let us find the ntiderivtive: x cos(x 2 ) dx. Tht is, for wht function will the Derivtive Rules produce x cos(x 2 )? We notice n inside function g(x) x 2, nd fctor x which is very close to the derivtive g (x) 2x. In fct, we cn get the exct derivtive of the inside function if we multiply the fctors by 2 nd 2: x cos(x 2 ) 2 cos(x2 ) (2x). This is just the kind of derivtive function produced by the Chin Rule: f(g(x)) f (g(x)) g (x) f (x 2 ) (2x)?? 2 cos(x2 ) (2x). We still need to find the outside function f. To remind us of the originl inside function, we write f(u), where the new vrible u represents u g(x) x 2. We must get f (u) 2 cos(u), n esy ntiderivtive: 2 cos(u) du f(u) + C 2 sin(u) + C. Now we restore the originl inside function to get our finl nswer: 2 cos(u) du 2 sin(u) + C 2 sin(x2 ) + C. The Chin Rule in Leibnitz nottion ( 2.5) reverses nd checks the bove computtion. Writing y 2 sin(u) nd u x2 : dy dx dy du du dx d ( du 2 sin(u)) d ( x 2 ) dx 2 cos(u) (2x) 2 cos(x2 ) (2x) x cos(x 2 ). Substitution Method. Given n ntiderivtive h(x) dx, try to find n inside function g(x) such tht g (x) is fctor of the integrnd: h(x) f(g(x)) g (x). This will often involve multiplying nd dividing by constnt to get the exct derivtive g (x). After fctoring out g (x), sometimes the remining fctor needs to be mnipulted to write it s function of u g(x).
2 2. Using the symbolic nottion u g(x), du du dx dx g (x) dx, write: h(x) dx f(g(x)) g (x) dx f(u) du, nd find the ntiderivtive f(u) du F (u) + C by whtever method. 3. Restore the originl inside function: h(x) dx f(u) du F (u) + C F (g(x)) + C. Exmples (3x+4) 3x+4 dx. The inside function is clerly u 3x+4, du 3 dx, so: (3x+4) 3x+4 dx 3 (3x+4) 3x+4 3 dx 3 u u du 3 u 3/2 du u5/2 + C 2 5 (3x+4)5/2 + C. x 3x+4 dx. Agin u 3x+4, so 3x+4 becomes u, but we must still express the remining fctor x in terms of u. We solve u 3x+4 to obtin x 3 u 4 3 : tht is, x 3 (3x+4) 4 3 : x 3x+4 dx 3 ( 3 (3x+4) 4 3 ) 3x+4 3 dx 3 ( 3 u 4 3 ) u du 9 u3/2 4 9 u/2 du u5/ u3/2 +C 2 45 (3x+4)5/ (3x+4)3/2 +C. sec 2 ( x) dx. We tke u x x /2, du x 2 x /2 dx 2 x dx sec 2 ( x) x dx 2 sec 2 ( x) 2 x dx sec 2 (u) du tn(u) + C tn( x) + C. Here we use the trig integrls from 3.9. sin(x) dx. We cnnot tke the inside function u sin(x), becuse ( + cos(x)) 2 its derivtive cos(x) is not fctor of the integrnd. We could tke u cos(x), but the best choice is u + cos(x), du sin(x) dx: sin(x) ( + cos(x)) 2 dx ( sin(x)) dx ( + cos(x)) 2 u 2 du u + C + cos(x) + C.
3 x dx. Tke u + x, du + x 2 x, so x u, x 2 u. x + x dx x (2 x) + x 2 x dx 2 2 u u 2 3u+2 2(u ) du 2 du u u u 3/2 3u /2 + 2u /2 du 2( 2 5 u5/2 2u 3/2 + 4u /2 ) + C 4 5 (+ x) 5/2 + 4(+ x) 3/2 8(+ x) /2 + C. Whew! Here we did not hve the derivtive fctor du dx 2 lredy present: x we hd to multipy nd divide by it to get du, then express the remining fctors in terms of u. By luck, the resulting du ws doble. sec 2 (x) tn(x) dx. Here we could tke u tn(x), du sec 2 (x) dx: sec 2 (x) tn(x) dx tn(x) sec 2 (x) dx u du 2 u2 + C 2 tn2 (x) + C. Alterntively, use the inside function z sec(x), dz tn(x) sec(x) dx: sec 2 (x) tn(x) dx sec(x) tn(x) sec(x) dx z dz 2 z2 + C 2 sec2 (x) + C. Thus 2 tn2 (x) nd 2 sec2 (x) re two different ntiderivtives, so wht bout the Antiderivtive Uniqueness Theorem ( 3.9)? In fct, the identity tn 2 (x)+ sec 2 (x) implies: 2 tn2 (x) sec2 (x). These give the sme ntiderivte fmily: 2 tn2 (x) + C 2 sec2 (x) + C!
4 Substitution for definite integrls. We hve, for u g(x): b f(g(x)) g (x) dx g(b) g() f(u) du. exmple: 3 2 x(+x2 ) 5 dx. Tking u +x 2, du 2x dx: 4 3 x(+x 2 ) 5 dx (+x2 ) 5 2x dx 2 u6 u7 u u5 du Integrl Symmetry Theorem: If f(x) is n odd function, mening f( x) f(x), then f(x) dx. Proof. By the Integrl Splitting Rule ( 4.2), we hve: f(x) dx f(x) dx + f(x) dx. Substituting u x, du ( ) dx in the first term, including in the limits of integrtion, nd using f( x) f(x), we get: () f(x) dx f(u) du f(x) ( ) dx f(u) du f( x) ( ) dx f(u) du f(x) dx. The lst equlity holds becuse the vrible of integrtion is merely suggestive, nd cn be chnged rbitrrily. Therefore f(x) dx + f(x) dx f(x) dx + f(x) dx, s desired. exmple: Evlute the definite integrl π π x cos(x) dx. Here substitution will not work, nd it is difficult to find n ntiderivtive. But since ( x) cos( x) (x cos(x)), the Theorem tells us the integrl must be zero. Geometriclly, the integrl is the signed re between the grph nd the xxis:
5 Since the function f(x) x cos(x) is odd, the grph hs rottionl symmetry round the origin, nd ech negtive re below the xxis cncels positive re bove the xxis.
F (x) dx = F (x)+c = u + C = du,
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