## Example 7.2.6(d): Applying Integration by Parts

As in the previous example, it seems that integration by parts does not apply because there is no product. But we can write*sin*and define the functions

^{5}(x) = sin(x) sin^{4}(x)Then

g'(x) = sin(x)so thatg(x) = -cos(x)f(x) = sinso that^{4}(x)f'(x) = 4 sin(x)^{3}cos(x)

*G(x) = f(x) g(x) = -cos(x) sin*and

^{4}(x)It seems as if we are stuck. Not only did we get the original integralsin^{5}(x) dx = -cos(x) sin^{4}(x) + 4 sin(x)^{3}cos^{2}(x) dx =

= -cos(x) sin^{4}(x) + 4 sin(x)^{3}(1 - sin^{2}(x)) dx =

= -cos(x) sin^{4}(x) + 4 sin(x)^{3}dx - 4 sin^{5}(x) dx

*sin*back, but we introduced

^{5}(x) dx*sin*which seems just about as complicated as the original integral. However, with a little more abstraction we can solve the original problem "by magic".

^{3}(x) dxLet

Define the functionsI_{k}= sin^{k}(x) dx = sin(x) sin^{k-1}(x) dx

Then

g'(x) = sin(x)so thatg(x) = -cos(x)f(x) = sinso that^{k-1}(x)f'(x) = (k-1) cos(x) sin^{k-2}(x)

Solving this expression forI_{k}= -cos(x) sin^{k-1}(x) + (k-1) cos^{2}sin^{k-2}(x) dx =

= -cos(x) sin^{k-1}(x) + (k-1) (1 - sin^{2}) sin^{k-2}(x) dx =

= -cos(x) sin^{k-1}(x) + (k-1) sin^{k-2}(x) dx - (k-1) sin^{k}(x) dx =

= -cos(x) sin^{k-1}(x) + (k-1) I_{k-2}- (k-1) I_{k}

*I*we get

_{k}But this is a recursive formula for computingI_{k}= -1/k cos(x) sin^{k-1}(x) + (k-1)/k I_{k-2}

*I*for any integer

_{k}*k*. In particular we have

Indeed, a - cumbersome - check would show that

I_{0}= sin^{0}(x) dx = 1 dx = xI_{1}= sin^{1}(x) dx = -cos(x)I_{2}= -1/2 cos(x) sin(x) + 1/2 I_{0}= -1/2 cos(x) sin(x) + 1/2 xI_{3}= -1/3 cos(x) sin^{2}(x) + 2/3 I_{1}= -1/3 cos(x) sin^{2}(x) - 2/3 cos(x)I_{4}= -1/4 cos(x) sin^{3}(x) + 3/4 I_{2}=

= -1/4 cos(x) sin^{3}+ 3/4 (-1/2 cos(x) sin(x) + 1/2 x)I_{5}= -1/5 cos(x) sin^{4}(x) + 4/5 I_{3}=

= -1/5 cos(x) sin^{4}(x) + 4/5 (-1/3 cos(x) sin^{2}(x) - 2/3 cos(x))

*I*, so that our recursive formula and the resulting answer for our particular example seem correct.

_{5}= sin^{5}