Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 17

The original PDF places the $-1$ outside the square root. Use the substitution rule with $t=\sqrt{x}$, so $dx=2tdt$ and the endpoints become $2$ and $3$. Polynomial division givesHence the definite integral is

Use the substitution rule $t=\tan x$, with $dt=(1+{\tan }^{2}x)dx$. The endpoints become $1$ and $\sqrt{3}$, so

Take $t=e^x>0$ in the substitution rule, so $dx=dt/t$. The transformed integral isThe two terms follow from the derivatives of the natural logarithm and the inverse tangent. Returning to $x$ gives the antiderivative

On an interval where $\tan x$ is defined, use the substitution rule $t=\tan x$. Then $dx=dt/(1+t^2)$ and ${\cos }^{2}x=1/(1+t^2)$, givingThe inverse tangent expression is valid on each interval $(j\pi -\pi /2,j\pi +\pi /2)$ separately. The original integrand is continuous for all real $x$, since its denominator is at least $1$. To obtain one continuous antiderivative on the whole real axis, add $j\pi /2$ on the $j$th such interval and use the continuous limits at their endpoints.