AI HL · AHL 3
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Integration by substitution
AHL 3.11
Screenshot and video to come
Screenshot and video to come
Use \(u=4x\) to find \(\int e^{4x}\,dx\).
Have a go first. Then open the working one line at a time.
\(u=4x\), \(du=4\,dx\), \(dx=\frac{du}{4}\)
The 4 comes from differentiating u.
\(\int e^u\cdot\frac{du}{4}=\frac14\int e^u\,du\)
Take the number outside.
\(\frac14e^u+c\)
The integral of \(e^u\) is \(e^u\).
\(\frac14e^{4x}+c\)
Replace u.
Find \(\int\frac{1}{5-2x}\,dx\). Use \(u=5-2x\).
Have a go first. Then open the working one line at a time.
\(u=5-2x\), \(du=-2\,dx\), \(dx=-\frac{du}{2}\)
The derivative of \(5-2x\) is \(-2\).
\(\int\frac1u\cdot\left(-\frac{du}{2}\right)=-\frac12\int\frac1u\,du\)
Take the number outside.
\(-\frac12\ln|u|+c\)
The integral of \(\frac1u\) is \(\ln|u|\).
\(-\frac12\ln|5-2x|+c\)
Replace u.
Evaluate \(\int_0^1x^2e^{x^3}\,dx\).
Have a go first. Then open the working one line at a time.
\(u=x^3\), \(du=3x^2\,dx\)
\(x^3\) is the inside of \(e^{x^3}\), and \(x^2\) is nearly its derivative.
\(\int x^2e^u\frac{du}{3x^2}=\frac13\int e^u\,du\)
The \(x^2\) cancels. No x is left.
\(x=0\Rightarrow u=0\); \(x=1\Rightarrow u=1\)
Put each limit into \(u=x^3\).
\(\frac13\left[e^u\right]_0^1=\frac{e-1}{3}=0.573\)
Check with the calculator.
Evaluate \(\int_0^1\frac{e^x}{3+e^x}\,dx\) using \(u=3+e^x\).
Have a go first. Then open the working one line at a time.
\(du=e^x\,dx\)
The numerator is exactly du.
\(\int\frac1u\,du=\ln u\)
u is positive, so no modulus is needed.
\(x=0\Rightarrow u=4\); \(x=1\Rightarrow u=3+e\)
Put the limits into \(u=3+e^x\).
\(\ln(3+e)-\ln 4=0.357\)
Top minus bottom.
Evaluate \(\int_0^1x\sqrt{x^2+3}\,dx\) using \(u=x^2+3\).
Have a go first. Then open the working one line at a time.
\(du=2x\,dx\), so \(x\,dx=\frac12du\)
The x in the integrand goes with the \(dx\).
\(\int\sqrt u\cdot\frac12du=\frac12\int u^{1/2}du\)
No x is left.
\(x=0\Rightarrow u=3\); \(x=1\Rightarrow u=4\)
Put the limits into \(u=x^2+3\).
\(\frac13\left[u^{3/2}\right]_3^4=\frac13(8-3\sqrt3)=0.935\)
Check with the calculator.
Evaluate \(\int_1^4\frac{e^{\sqrt x}}{\sqrt x}\,dx\) using \(u=\sqrt x\).
Have a go first. Then open the working one line at a time.
\(u=x^{1/2}\), \(du=\frac{1}{2\sqrt x}dx\), so \(\frac{dx}{\sqrt x}=2\,du\)
The \(\frac1{\sqrt x}\) in the integrand uses up the derivative.
\(\int e^u\cdot2\,du=2e^u\)
No x is left.
\(x=1\Rightarrow u=1\); \(x=4\Rightarrow u=2\)
\(\sqrt1=1\), \(\sqrt4=2\).
\(2(e^2-e)=9.34\)
Check with the calculator.
Evaluate \(\int_0^{\pi/12}\tan 4x\,dx\).
Have a go first. Then open the working one line at a time.
\(\tan4x=\frac{\sin4x}{\cos4x}\)
The numerator is a multiple of the derivative of the denominator.
\(u=\cos4x\), \(du=-4\sin4x\,dx\)
So \(\sin4x\,dx=-\frac14du\).
\(-\frac14\int\frac1u\,du\); \(x=0\Rightarrow u=1\), \(x=\frac{\pi}{12}\Rightarrow u=\frac12\)
\(\cos\frac{\pi}{3}=\frac12\).
\(-\frac14\left[\ln u\right]_1^{1/2}=-\frac14\ln\frac12=\frac14\ln2=0.173\)
A positive value: the area under \(\tan4x\) is above the axis.
Use \(u=x+4\) to find \(\int_0^1\sqrt{x^3+4x^2}\,dx\).
Have a go first. Then open the working one line at a time.
\(\sqrt{x^3+4x^2}=\sqrt{x^2(x+4)}=x\sqrt{x+4}\)
This is true for \(x>0\).
\(x=u-4\), \(dx=du\), so \(\int(u-4)\sqrt u\,du=\int\left(u^{3/2}-4u^{1/2}\right)du\)
Every x becomes \(u-4\).
\(x=0\Rightarrow u=4\); \(x=1\Rightarrow u=5\)
Put the limits into \(u=x+4\).
\(\left[\frac25u^{5/2}-\frac83u^{3/2}\right]_4^5=1.08\)
Check with the calculator.
Use \(x=\sin u\) to evaluate \(\int_0^{1/2}\frac{1}{\sqrt{1-x^2}}\,dx\). (A check only: this type is not examined.)
Have a go first. Then open the working one line at a time.
\(x=\sin u\), \(dx=\cos u\,du\)
Here x is written in terms of u.
\(\sqrt{1-\sin^2u}=\cos u\), so the integrand becomes \(\frac{\cos u}{\cos u}du=du\)
\(1-\sin^2u=\cos^2u\).
\(x=0\Rightarrow u=0\); \(x=\frac12\Rightarrow u=\frac{\pi}{6}\)
\(\sin\frac{\pi}{6}=\frac12\).
\(\left[u\right]_0^{\pi/6}=\frac{\pi}{6}=0.524\)
So \(\int\frac{1}{\sqrt{1-x^2}}dx=\arcsin x+c\).