turning the big gun around
In the last few sections, we have learned how to evaluate an indefinite integral. And the evaluation of definite integrals follows. We have learned how to use trigonometry, algebra, and anything we have to throw at that stupid elongated ‘S’.
We have learned several things—differentiation formulas can sometimes be turned around to attack integration problems. Here, we learn how to use the Product Rule to solve a broader range of integration problems.
If you still remember, the Product Rule states—if h(x)=f(x)*g(x), then h’(x)=f’(x)g(x)+f(x)g’(x). Somebody realized this could be a perfect weapon for integration. Think about this: integrate xsin(x). We feel that this might be one of the two terms on the right hand side of a Product Rule application. Suppose sin(x)=f’(x) and x=g(x). Now, f’(x)g(x)=h’(x)f(x)g’(x).
And now we integrate. The integral of the left hand side (what we are trying to find) must be equal to the integral of the right hand side, which is h(x) minus the integral of f(x)g’(x) plus some constant. So, we have transferred the problem into finding the integral of f(x)g’(x). Now you are asking: so what? We have traded one problem for another!
That’s where the tricks come in. When you apply this procedure, oh, it’s called integration by parts, you need to make sure that evaluating the integral of f(x)g’(x) is easier than the original problem. Up there, we set sin(x)=f’(x) and x=g(x). Now, f(x)g’(x) is just –cos(x). You should be able to know the integral immediately. Down below is a table that will help:
We have learned several things—differentiation formulas can sometimes be turned around to attack integration problems. Here, we learn how to use the Product Rule to solve a broader range of integration problems.
If you still remember, the Product Rule states—if h(x)=f(x)*g(x), then h’(x)=f’(x)g(x)+f(x)g’(x). Somebody realized this could be a perfect weapon for integration. Think about this: integrate xsin(x). We feel that this might be one of the two terms on the right hand side of a Product Rule application. Suppose sin(x)=f’(x) and x=g(x). Now, f’(x)g(x)=h’(x)f(x)g’(x).
And now we integrate. The integral of the left hand side (what we are trying to find) must be equal to the integral of the right hand side, which is h(x) minus the integral of f(x)g’(x) plus some constant. So, we have transferred the problem into finding the integral of f(x)g’(x). Now you are asking: so what? We have traded one problem for another!
That’s where the tricks come in. When you apply this procedure, oh, it’s called integration by parts, you need to make sure that evaluating the integral of f(x)g’(x) is easier than the original problem. Up there, we set sin(x)=f’(x) and x=g(x). Now, f(x)g’(x) is just –cos(x). You should be able to know the integral immediately. Down below is a table that will help:
f'(x)=?
f(x)=?

g(x)=?
g'(x)=?

after trying out for the best f(x) and g'(x), bam! solve the puzzle. and take a break.