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How do I solve an integration by substitution problem?

I think it’s best I work through an example with you as these problems can vary quite a lot, but the general methods used are the same.

Example: Use the substitution u=x2+5 to find: The integral of (x3/sqrt(x2+5)).dx between the limits of 2 and 1.

So the whole idea of using a substitution here is to simplify the integration for us. The first thing we must do is substitute the given substitution in, otherwise there wouldn’t be much point! In doing this, we also need to replace the .dx in the integration, we do this by finding du/dx and then rearranging for dx; in this particular example, du/dx = 2x, so dx = (1/2x)du, so the integral becomes (x3/sqrt(x2+5)). (1/2x).du =( x2/2sqrt(x2+5)).du. We then use the substitution to get the x’s in terms of u: the numerator, x2 becomes u-5 (as u=x2+5), the denominator, 2sqrt(x2+5) becomes 2sqrt(u). Finally, modifying the limits in terms of u: since the top limit is x=2, then this is equivalent to u=22+5=9, and the bottom limit becomes 12+5=6.

We now have our rephrased integration problem: Integrate ((u-5)/2sqrt(u)).du between the limits of 9 and 6. Notice we can split up the integrand (the thing we’re integrating): =(u/2sqrt(u))-(5/2sqrt(u)). You know that this is equivalent to 0.5u1/2-5.5u-1/2, which when integrated is (1/3)u3/2-5u1/2. The only thing left to do now is apply the limits: [(1/3)(93/2)-5(91/2)]-[ [(1/3)(63/2)-5(61/2)] = 9-15-2sqrt(6)+5sqrt(6) = 7sqrt(6)-6

The steps to solving other substitution problems are very similar to the ones detailed above, obviously the manipulations will be different, but the ideas are the same.

Andrew D. A Level Maths tutor, GCSE Maths tutor, A Level Further Math...

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