Recursive version of integration by parts

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General description of technique

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Examples

Sine-squared function

For further information, refer: Sine-squared function#Integration

There are many ways of integrating sin2. One of these uses the recursive version of integration by parts. This method is given below:


sin2xdx=(sinx)(cosx)(cosx)(cosx)dx=sinxcosx+cos2xdx

We now rewrite cos2x=1sin2x and obtain:

sin2xdx=sinxcosx+(1sin2x)dx

Setting I to be a choice of antiderivative so that the above holds without any freely floating constants, we get:

I=sinxcosx+xI

Rearranging, we get:

2I=xsinxcosx

This gives:

I=xsinxcosx2

So the general antiderivative is:

xsinxcosx2+C


Secant-cubed function

For further information, refer: Secant-cubed function#Integration


We rewrite sec3x=secxsec2x and perform integration by parts, taking sec2 as the part to integrate. We use that an antiderivative of sec2x is tanx whereas the derivative of secx is secxtanx:

sec3xdx=secx(sec2x)dx=secxtanx(secxtanx)tanxdx=secxtanxsecxtan2xdx

We now use the fact that tan2x+1=sec2x, or more explicitly, tan2x=sec2x1, to rewrite this as:

sec3xdx=secxtanxsecx(sec2x1)dx=secxtanxxsec3xdx+secxd

We now use the integration of the secant function to simplify this as:

sec3xdx=secxtanxsec3xdx+ln|secx+tanx|

We can choose an antiderivative I of sec3 so that the above equality (between the left-most and right-most expression) holds without any additive constant adjustment, and we get:

I=secxtanxI+ln|secx+tanx|

We rearrange and obtain:

2I=secxtanx+ln|secx+tanx|

Dividing by 2, we get:

I=secxtanx+ln|secx+tanx|2

The general antiderivative expression is thus:

secxtanx+ln|secx+tanx|2+C