Question 8
Use the substitution $u=\tan(x)$ to determine $\displaystyle\int\tan(x)\sec^2(x)\,dx$.
Not quite. The $\tfrac12$ is right, but $\int u\,du=\tfrac12u^2$, so the power of $u$ has to go up to $2$ as well. Differentiating $\tfrac12\tan(x)$ gives $\tfrac12\sec^2(x)$, which has lost the $\tan(x)$. In 2022, 5.3% of Queensland students chose (A).
That’s it. Step through the hints below to see why. In 2022, 80.8% of Queensland students chose (B).
Not quite. $\tan(x)$ is $u$ itself, so this treats $\int u\,du$ as if it were $u$. Differentiating $\tan(x)$ gives $\sec^2(x)$, with no $\tan(x)$ factor. In 2022, 5.7% of Queensland students chose (C).
Not quite. The power went up to $2$, but you also divide by the new power. Differentiating $\tan^2(x)$ gives $2\tan(x)\sec^2(x)$, twice the integrand. In 2022, 7.8% of Queensland students chose (D).
An antiderivative is right when its derivative gives back the integrand. Pick an option to differentiate it. The second readout compares the dashed curve with the blue one at one value of .
The dashed curve is $\tfrac12\sec^2(x)$. It crosses the blue curve once, then falls well below it. The factor $\tan(x)$ has gone missing because $u$ was never squared.
The dashed curve sits exactly on the blue one at every $x$. This antiderivative works.
The dashed curve is $\sec^2(x)$. It starts at $1$ when the blue curve starts at $0$, so the factor $\tan(x)$ has been lost.
The dashed curve is exactly twice the blue one. The power went up to $2$ but nothing divided by $2$.
Differentiate the substitution to see what $du$ replaces.
$$u=\tan(x)\ \Rightarrow\ \frac{du}{dx}=\sec^2(x)$$
$$du=\sec^2(x)\,dx$$
Swap each piece. The $\tan(x)$ becomes $u$, and $\sec^2(x)\,dx$ becomes $du$. No $x$ is left over.
$$\int\tan(x)\,\underbrace{\sec^2(x)\,dx}_{du}=\int u\,du$$
Integrate with the power rule, then substitute back.
$$\int u\,du=\frac{u^2}{2}+c=\frac12\tan^2(x)+c$$
Check with the chain rule: the derivative of $\tfrac12\tan^2(x)$ is $\tfrac12\cdot2\tan(x)\cdot\sec^2(x)$. If a friend gets $\tfrac12\sec^2(x)+c$, that is also right. Since $\sec^2(x)=1+\tan^2(x)$, the two answers differ only by a constant.
With $u=\tan(x)$, $du=\sec^2(x)\,dx$.
$$\int\tan(x)\sec^2(x)\,dx=\int u\,du$$
$$=\frac{u^2}{2}+c=\frac12\tan^2(x)+c$$
How Queensland students answered in 2022
From the QCAA Specialist Mathematics subject report for 2022. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2022 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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