Integration using a given substitution | CAIE 9709 Pure Mathematics 3 Notes

Chapters

Pure Mathematics 3

Mathematics9709

Rewrite the whole integral in one variable

The no-stray-x audit

replace x

rewrite the integrand

replace dx

use du = u′ dx

handle limits

new limits or return to x

Never integrate a mixture of x and u.

A substitution is complete only when the whole integral uses one variable.

The letter $u$ names the new variable. The relation for $du$ tells you which factor, together with $dx$ , is replaced. For example, $du=2x\,dx$ means the whole product $2x\,dx$ becomes $du$ .

Differentiate the given relation to connect $du$ and $dx$ .
Replace every occurrence of $x$ and replace $dx$ . Stop if both $x$ and $u$ remain.
For a definite integral, convert the limits. For an indefinite integral, substitute back before the final answer.

Examiner note

Choosing a substitution is not required by the syllabus. Use the one supplied and show the derivative link explicitly.

See the cancellation before integrating

Worked example

Given u = x² + 1

\int2x(x^2+1)^5\,dx

u=x^2+1

du=2x\,dx

The factor $2x\,dx$ becomes exactly $du$ :

\int u^5\,du=\frac{u^6}{6}+C

Answer

\frac{(x^2+1)^6}{6}+C

Common mistake

Writing

u=x^2+1

does not change the integral by itself. The essential link is the relation for

du

and the complete rewrite in

u

(Pure Mathematics 2 & 3 Coursebook, Ch. 8.4)

Chain substitution and partial fractions

Worked example

Current-paper nine-mark chain

\int_{\ln3}^{\ln5}\frac{2e^{2x}}{e^{2x}-3e^x+2}\,dx

With $u=e^x$ , $du=e^x dx=u\,dx$ , so $dx=du/u$ . The limits become 3 and 5.

\int_3^5\frac{2u}{(u-1)(u-2)}\,du

\frac{2u}{(u-1)(u-2)}

=-\frac2{u-1}+\frac4{u-2}

I_1=\left[-2\ln(u-1)\right]_3^5

I_2=\left[4\ln(u-2)\right]_3^5

I=I_1+I_2

Answer

\ln\!\left(\frac{81}{4}\right)

(9709/32/O/N/24 Q11(b))

Your turnIndependent check [5 marks]

Using $u=e^x$ , evaluate $\displaystyle\int_0^{\ln2}\frac{e^x}{e^x+1}\,dx$ exactly.

Show worked answer

Since $du=e^x dx$ , the new limits are $u=1$ and $u=2$ .

\int_1^2\frac1{u+1}\,du

=\left[\ln(u+1)\right]_1^2

Answer

\ln\!\left(\frac32\right)

Rewrite the whole integral in one variable

The no-stray-x audit

replace x

rewrite the integrand

replace dx

use du = u′ dx

handle limits

new limits or return to x

Never integrate a mixture of x and u.

A substitution is complete only when the whole integral uses one variable.

The letter $u$ names the new variable. The relation for $du$ tells you which factor, together with $dx$ , is replaced. For example, $du=2x\,dx$ means the whole product $2x\,dx$ becomes $du$ .

Differentiate the given relation to connect $du$ and $dx$ .
Replace every occurrence of $x$ and replace $dx$ . Stop if both $x$ and $u$ remain.
For a definite integral, convert the limits. For an indefinite integral, substitute back before the final answer.

Examiner note

Choosing a substitution is not required by the syllabus. Use the one supplied and show the derivative link explicitly.

See the cancellation before integrating

Worked example

Given u = x² + 1

\int2x(x^2+1)^5\,dx

u=x^2+1

du=2x\,dx

The factor $2x\,dx$ becomes exactly $du$ :

\int u^5\,du=\frac{u^6}{6}+C

Answer

\frac{(x^2+1)^6}{6}+C

Common mistake

Writing

u=x^2+1

does not change the integral by itself. The essential link is the relation for

du

and the complete rewrite in

u

(Pure Mathematics 2 & 3 Coursebook, Ch. 8.4)

Chain substitution and partial fractions

Worked example

Current-paper nine-mark chain

\int_{\ln3}^{\ln5}\frac{2e^{2x}}{e^{2x}-3e^x+2}\,dx

With $u=e^x$ , $du=e^x dx=u\,dx$ , so $dx=du/u$ . The limits become 3 and 5.

\int_3^5\frac{2u}{(u-1)(u-2)}\,du

\frac{2u}{(u-1)(u-2)}

=-\frac2{u-1}+\frac4{u-2}

I_1=\left[-2\ln(u-1)\right]_3^5

I_2=\left[4\ln(u-2)\right]_3^5

I=I_1+I_2

Answer

\ln\!\left(\frac{81}{4}\right)

(9709/32/O/N/24 Q11(b))

Your turnIndependent check [5 marks]

Using $u=e^x$ , evaluate $\displaystyle\int_0^{\ln2}\frac{e^x}{e^x+1}\,dx$ exactly.

Show worked answer

Since $du=e^x dx$ , the new limits are $u=1$ and $u=2$ .

\int_1^2\frac1{u+1}\,du

=\left[\ln(u+1)\right]_1^2

Answer

\ln\!\left(\frac32\right)