Integration by Reverse Chain Rule

Chain Rule for Differentiation

\int \frac{f^{'} (x)}{f (x)} d x = ln | f (x) |

If integrating expressions of the form

\int k \frac{f^{'} (x)}{f (x)} d x

then try

ln | f (x) |

differentiate to check, and adjust with a constant if required.

\int f^{'} (x) {(f (x))}^{n} d x = {(f (x))}^{n + 1}

If integrating expressions of the form

\int k f^{'} (x) {(f (x))}^{n} d x

then try

{(f (x))}^{n + 1}

differentiate to check, and adjust with a constant if required.

Exercise 11D page 302 [ MS (Archive)]

Ex 11D Qu 1 (i): Given that ${sec}^{2} x$ is the derivative of $tan x$ then the trial solution should be $y = {tan}^{3} x$ $\Rightarrow \frac{d y}{d x} = 3 {sec}^{2} x {tan}^{2} x$ So, the trial solution needs to be adjusted by $\frac{1}{3}$ $\int {sec}^{2} x {tan}^{2} x d x = \frac{1}{3} {tan}^{3} x + C$

Ex 11D Qu 1 (j): Firstly, expand the integrand to give $\int {sec}^{2} x + {sec}^{2} x {tan}^{2} x d x$ Then integrate term-wise, the first term from standard integrals and the second from part (i) of this question. Giving $\int {sec}^{2} x (1 + {tan}^{2} x) d x = tan x + \frac{1}{3} {tan}^{3} x + C$