H5 - Integration by substitution

Recognising the reverse chain rule

Can you integrate $6x(3x^2+4)^3$ without expanding the cube? Differentiating $(3x^2+4)^4$ produces $4(3x^2+4)^3(6x)$. The integral is therefore one quarter of that fourth power. The outside factor $6x$ is the clue: it is the derivative of the inside expression.

A substitution names an expression, such as $u=3x^2+4$, as a new variable. Differentiating gives $du=6x\,dx$. This is a compact way of reversing the chain rule: the differential changes along with the expression.

The excluded power $-1$ needs a logarithm instead. The booklet supplies $\int f'(x)/f(x)\,dx=\ln|f(x)|+C$, on an interval where $f(x)\ne0$. A constant multiple of the inside derivative can be adjusted outside the integral.

Now change only the outer power, from $3$ to $-1$. The same substitution works, but the antiderivative changes from a power to a logarithm.

You can recognise the pattern without writing a substitution every time. Writing it remains a useful way to check the missing constant factor.

Change the outer power once more in your turn. The inside derivative still matches exactly.

Changing the whole integrand

Sometimes the inside derivative does not remove every $x$, but a little rearrangement does. In $x\sqrt{x+2}$, naming $x+2$ as $u$ also tells you that $x=u-2$. That extra line turns the product into ordinary powers.

Choose a substitution that leads directly to something you can integrate. Before integrating, every expression and the differential must use the new variable. An integral containing both $x$ and $u$ is unfinished.

Change only the constant inside the root, from $2$ to $3$. The rearranged factor and the original domain both change with it.

There is a second neat route through Worked example 3: set $w=\sqrt{x+2}$. Then $x=w^2-2$ and $dx=2w\,dw$, giving $\int(2w^4-4w^2)\,dw=2w^5/5-4w^3/3+C$. Naming the inside or naming its root leads to exactly the same answer after substituting back. Both routes use just one substitution. These power antiderivatives are not in the formula booklet: learn the power rule.

The next question changes the shift again. State the differential as well as the substitution if you use this route.

Substitution in the other direction and domains

A substitution can instead give the original variable in terms of a new one. With $x=\sin\theta$, the expression $1-x^2$ becomes $\cos^2\theta$. This is useful only if you also handle the square root's sign.