Question 1
Use the substitution $u=\cos(x)$ to determine $\displaystyle\int\sin(x)\cos^{3}(x)\,dx$.
Not quite. You dropped the negative sign in $du=-\sin(x)\,dx$, so you wrote $\int u^{3}\,du=\dfrac{u^{4}}{4}+c$.
Not quite. You handled the sign correctly, but you antidifferentiated $u^{3}$ as $u^{4}$ and forgot to divide by 4.
That’s it. Step through the hints below to see why.
Not quite. You never converted $dx$ to $du$. You left the $\sin(x)$ outside as if it were a constant and only antidifferentiated $\cos^{3}(x)=u^{3}$. That gives $\sin(x)\cdot\dfrac{u^{4}}{4}=\dfrac{1}{4}\sin(x)\cos^{4}(x)$.
Differentiate the substitution, then rearrange to isolate the product $\sin(x)\,dx$.
$$\begin{aligned}u&=\cos(x)\\\frac{du}{dx}&=-\sin(x)\\\sin(x)\,dx&=-du\end{aligned}$$
To change the variable, you need to know what $dx$ becomes in terms of $du$. The integrand contains exactly $\sin(x)\,dx$, so you isolate that product to replace it in one go.
Rewrite the integrand: $\cos^{3}(x)=u^{3}$ and $\sin(x)\,dx=-du$.
$$\begin{aligned}\int\sin(x)\cos^{3}(x)\,dx&=\int u^{3}\,(-du)\\&=-\int u^{3}\,du\end{aligned}$$
Once every $x$ (including the $dx$) is replaced, the integral is a plain power of $u$.
Antidifferentiate the power, then substitute back $u=\cos(x)$.
$$-\int u^{3}\,du=-\frac{u^{4}}{4}+c=-\frac{1}{4}\cos^{4}(x)+c$$
$\int u^{n}\,du=\dfrac{u^{n+1}}{n+1}+c$ with $n=3$: raise the power by 1 and divide by the new power, 4. The question is in terms of $x$, so the answer must be too.
Check by differentiating. You should get the integrand back.
$$\begin{aligned}&\frac{d}{dx}\left[-\frac{1}{4}\cos^{4}(x)\right]\\&=-\frac{1}{4}\cdot4\cos^{3}(x)\cdot\left(-\sin(x)\right)\\&=\sin(x)\cos^{3}(x)\end{aligned}$$
Differentiating is the fastest way to confirm an antiderivative, and it catches sign errors.
With $u=\cos(x)$ and $\sin(x)\,dx=-du$, the integral becomes $-\int u^{3}\,du=-\dfrac{u^{4}}{4}+c$. Substituting back gives $-\dfrac{1}{4}\cos^{4}(x)+c$.
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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