Change both the variable and its differential · source reference

Printed question

(a) Find $\int\frac1{3-4t^2}dt$. [3]

(b) Use the substitution $u=5^x$ to find $\int5^x\cos^2(5^x)dx$. [5]

[2010/HCI/Prelim/I/6]

Printed working

(a)

$$\int\frac1{3-4t^2}dt=-\frac14\int\frac1{t^2-3/4}dt=-\frac14\int\frac1{t^2-(\sqrt3/2)^2}dt=-\frac14\frac1{2(\sqrt3/2)}\ln\left|\frac{t-\sqrt3/2}{t+\sqrt3/2}\right|+c=-\frac{\sqrt3}{12}\ln\left|\frac{2t-\sqrt3}{2t+\sqrt3}\right|+c.$$

(b)

$u=5^x\Rightarrow du/dx=5^x\ln5$, so $dx/du=1/(5^x\ln5)=1/(u\ln5)$.

$$\int5^x\cos^2(5^x)dx=\int u\cos^2u\frac1{u\ln5}du=\frac1{\ln5}\int\cos^2u\,du=\frac1{2\ln5}\int(1+\cos2u)du=\frac1{2\ln5}\left[u+\frac{\sin2u}2\right]+c=\frac1{2\ln5}\left[5^x+\frac{\sin(2(5^x))}2\right]+c.$$

Teacher notes

(a) MF27 $\int dx/(x^2-a^2)=\frac1{2a}\ln|(x-a)/(x+a)|+c$, replacing $x$ by $t$, $a$ by $\sqrt3/2$. Put in the modulus sign in $\ln|\cdots|$.

(b) Integration by substitution: $\int f(x)dx=\int f(x)\frac{dx}{du}du=\int g(u)\frac{dx}{du}du$. Express the integrand entirely in $u$. For indefinite integrals in $x$, the final answer must also be in terms of $x$.

Note: $\int g(u)(dx/du)du\ne(\int g(u)du)(dx/du)$. Example: $\int2u\cdot u\,du\ne u\int2u\,du$ since the first is $\int2u^2du$, whereas the second has $u$ outside the integral.

Printed answer key

(a) $-\frac{\sqrt3}{12}\ln|(2t-\sqrt3)/(2t+\sqrt3)|+c$; (b) $\frac1{2\ln5}[5^x+\sin(2(5^x))/2]+c$.