Every differentiation rule for $e^x$, $\ln x$ and the trig functions reverses into an integration rule — learn the results, and combine them term by term.
8 min read4 worked examples
The standard results
So far, integration has meant reversing the power rule on polynomials. Year 2 introduces derivatives of $e^x$, $\ln x$ and the trig functions — and each of those differentiation results reverses directly into an integration result, with no extra technique required.
The modulus in $\ln|x|$ matters: $\ln x$ is only defined for $x>0$, but $\dfrac1x$ is a perfectly good function for $x<0$ too, and $\ln|x|$ is the antiderivative that works on both sides of the origin. Also easy to lose a sign: differentiating $-\cos x$ gives $+\sin x$, so it's $-\cos x$ (not $\cos x$) that integrates $\sin x$.
Combining terms
Just as with $x^n$, a sum or difference of standard functions is integrated term by term, and constant multiples carry straight through. There's no new machinery here — only more functions in the toolkit alongside $x^n$.
Integrate the polynomial term with the power rule, and the 1/x term with the standard result.
▪ solved
Look for this pattern
Whenever an integrand is a fraction with a polynomial numerator and a monomial denominator, dividing through term by term is usually the fastest route to a standard-form integral.
Worked example 4
Exact value with a negative result · Challenge
Find the exact value of $\displaystyle\int_{\pi/6}^{\pi/2} \left(3\cos x - 4\sin x\right)dx$.
MethodIntegrate, then evaluate carefully — the answer needn't be positive
1
$\displaystyle\int \left(3\cos x - 4\sin x\right)dx = 3\sin x + 4\cos x$.
Apply the standard sin and cos results; note the sign on the cos term flips because ∫sin x dx = −cos x.
2
At $x=\dfrac\pi2$: $3\sin\dfrac\pi2+4\cos\dfrac\pi2 = 3(1)+4(0)=3$.
Substitute the upper limit.
3
At $x=\dfrac\pi6$: $3\sin\dfrac\pi6+4\cos\dfrac\pi6 = 3\left(\dfrac12\right)+4\left(\dfrac{\sqrt3}2\right) = \dfrac32+2\sqrt3$.
Substitute the lower limit, using exact trig values.
Subtract; since 2√3 ≈ 3.46 > 1.5, the exact value is negative.
▪ solved
A negative definite integral is not an error
A definite integral gives the signed area between the curve and the x-axis. Where the curve dips below the axis more than it rises above it over the interval, the exact value comes out negative — that's a correct result, not a mistake.
Common mistake
Dropping the modulus signs from $\ln|x|$, or forgetting the minus sign that comes from integrating $\sin x$. Both errors are easy to make under time pressure — it's worth differentiating your answer back mentally as a quick check, since that should always return the original integrand.