So if we cannot compute the integral h x dx directly we often look for a new

So if we cannot compute the integral h x dx directly

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So, if we cannot compute the integral h ( x ) dx directly, we often look for a new variable u and function f ( u ) for which h ( x ) dx = f ( u ( x )) du dx dx = f ( u ) du , where the second integral is easier to evaluate than the first. NOTES In deciding how to choose a new variable, there are several things to look for: terms that are derivatives of other terms (or pieces thereof ) and terms that are particularly troublesome. (You can often substitute your troubles away.) EXAMPLE 6.2 Using Substitution to Evaluate an Integral Evaluate ( x 3 + 5) 100 (3 x 2 ) dx . Solution You probably cannot evaluate this as it stands. However, observe that d dx ( x 3 + 5) = 3 x 2 , which is part of the integrand. This leads us to make the substitution u = x 3 + 5, so that du = d dx ( x 3 + 5) dx = 3 x 2 dx . This gives us ( x 3 + 5) 100 u 100 (3 x 2 ) dx du = u 100 du = u 101 101 + c . We are not done quite yet. Since we invented the new variable u , we need to convert back to the original variable x , to obtain ( x 3 + 5) 100 (3 x 2 ) dx = u 101 101 + c = ( x 3 + 5) 101 101 + c . It’s always a good idea to perform a quick check on the antiderivative. (Remember that integration and differentiation are inverse processes!) Here, we compute d dx ( x 3 + 5) 101 101 = 101( x 3 + 5) 100 (3 x 2 ) 101 = ( x 3 + 5) 100 (3 x 2 ) , which is the original integrand. This confirms that we have indeed found an antiderivative.
4-53 SECTION 4.6 . . Integration by Substitution 395 INTEGRATION BY SUBSTITUTION Integration by substitution consists of the following general steps, as illustrated in example 6.2. Choose a new variable u : a common choice is the innermost expression or “inside” term of a composition of functions. (In example 6.2, note that x 3 + 5 is the inside term of ( x 3 + 5) 100 .) Compute du = du dx dx . Replace all terms in the original integrand with expressions involving u and du . Evaluate the resulting ( u ) integral. If you still can’t evaluate the integral, you may need to try a different choice of u . Replace each occurrence of u in the antiderivative with the corresponding expression in x . Always keep in mind that finding antiderivatives is the reverse process of finding derivatives. In example 6.3, we are not so fortunate as to have the exact derivative we want in the integrand. EXAMPLE 6.3 Using Substitution: A Power Function Inside a Cosine Evaluate x cos x 2 dx . Solution Notice that d dx x 2 = 2 x . Unfortunately, we don’t quite have a factor of 2 x in the integrand. This does not present a problem, though, as you can always push constants back and forth past an integral sign. Notice that we can rewrite the integral as x cos x 2 dx = 1 2 2 x cos x 2 dx . We now substitute u = x 2 , so that du = 2 x dx and we have x cos x 2 dx = 1 2 cos x 2 cos u (2 x ) dx du = 1 2 cos u du = 1 2 sin u + c = 1 2 sin x 2 + c . Again, as a check, observe that d dx 1 2 sin x 2 = 1 2 cos x 2 (2 x ) = x cos x 2 , which is the original integrand. EXAMPLE 6.4 Using Substitution: A Trigonometric Function Inside a Power Evaluate (3 sin x + 4) 5 cos x dx .

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