Integration by Parts

Recall the product rule: (fg)' = f'g + fg'. Now, if and then see the blackboard (or your text!) to see how the product rule produces the formula: which is the formula for integration by parts . It is crucial to realize that du = f ' ( x ) dx and dv = g'( x ) dx are the respective differentials.

As a very rough strategy for deciding whether or not to use integration by parts you should be alert for an integral (which is not an "easy" one in disguise) in which the integrand (that's the part behind the integral sign) can be expressed as a product udv where u is some function that becomes "simpler" (or at least not more complicated) when differentiated and dv is an expression which is "easy" (or at least possible ) to integrate. N.B. u and dv must be such that their product is the entire original integrand.

e.g. Find

> F:=Int(x*exp(1)^x,x);

This type of integral is not on the easy list yet . Notice that x gets simpler wjen differentiated and that is easy to integrate. Therefore let (so that ) and (so that ) and apply the formula:

Here's what happens if you chose the "right" u:

> intparts(F,x);

> value(%);

> factor(%);

Here's what happens if you chose the "wrong" u:

> intparts(F,exp(1)^x);

You can see that the part of the formula gets harder!

e.g. Find

> F:=Int(x^2*exp(1)^x,x);

> intparts(F,x^2);

You can see from the above that you can now finish off the integration by using the first result for or you can use integration by parts again on the unevaluated integral. Here's the answer:

> value(%);factor(%);

Find .

This always fools the unwary! Many newbies will write = which is WRONG (why?). This one is dne by parts using and :

> F:=Int(ln(x),x);

> intparts(F,ln(x));

> value(%);

More

Find

> Int(exp(1)^x*cos(x),x);

> G:=value(%);

Whoa!! MY answer is ! Are they the same?

> F:=1*exp(x)*(sin(x)+cos(x))/2;

> simplify(F-G);

Find

> F:=Int(arccos(w),w = 0 .. 1/2);

> intparts(F,arccos(w));

> value(%);evalf(%);

e.g. A Reduction Formula.

e.g. Use integration by parts to prove that for and use the formula to find and .

> g:=sin(x)^2:h:=sin(x)^3:

> Int(g,x)=int(g,x)+C;Int(h,x)=int(h,x)+K;