The t-Substitution
Some integrands — like $\dfrac{1}{2+\cos x}$ — resist every standard trig identity. The Weierstrass substitution $t = \tan(x/2)$ rewrites $\sin x$, $\cos x$ and $dx$ as rational functions of $t$, turning the problem into one a Year 12 student can solve with partial fractions or a standard arctan form. This lesson teaches you the three substitution formulas, the algebraic clean-up, and how to handle definite limits.
Recall the double-angle identities. Using $t = \tan(x/2)$ and a right-triangle, express $\sin x$ and $\cos x$ as rational functions of $t$. Before checking — what do you expect $dx$ in terms of $dt$ to look like? Sketch your reasoning below.
Every t-substitution problem rewards two habits: substitute all three pieces at once ($\sin x$, $\cos x$, $dx$), then simplify the algebra fully before integrating. The substitution always produces a rational function of $t$ — if you find anything else, an algebra slip has happened.
The substitute-simplify-integrate flow: (1) replace $\sin x$, $\cos x$ and $dx$ with their $t$-forms, (2) clear the $1+t^2$ factors and combine, (3) integrate the rational function (often a standard arctan form or partial fractions).
$\sin x = \dfrac{2t}{1+t^2}$ · $\cos x = \dfrac{1-t^2}{1+t^2}$ · $dx = \dfrac{2}{1+t^2}\,dt$
Key facts
- $t = \tan(x/2)$ is the Weierstrass substitution
- $\sin x = \dfrac{2t}{1+t^2}$, $\cos x = \dfrac{1-t^2}{1+t^2}$
- $dx = \dfrac{2}{1+t^2}\,dt$
- $\tan x = \dfrac{2t}{1-t^2}$ follows from sin/cos
Concepts
- Why $t = \tan(x/2)$ converts any rational function of $\sin x, \cos x$ into a rational function of $t$
- Why $1+t^2$ factors are the key simplifier (they come from $\sec^2(x/2)$)
- How definite limits transform under the substitution
Skills
- Apply $t = \tan(x/2)$ to integrals of the form $\int R(\sin x, \cos x)\,dx$
- Simplify to a rational function of $t$ and integrate (arctan or partial fractions)
- Handle definite integrals by transforming limits
Let $t = \tan(x/2)$. Then by the right-triangle (opposite $t$, adjacent $1$, hypotenuse $\sqrt{1+t^2}$):
$\sin(x/2) = \dfrac{t}{\sqrt{1+t^2}}$ and $\cos(x/2) = \dfrac{1}{\sqrt{1+t^2}}$.
Double-angle for $\sin x$:
$\sin x = 2\sin(x/2)\cos(x/2) = 2 \cdot \dfrac{t}{\sqrt{1+t^2}} \cdot \dfrac{1}{\sqrt{1+t^2}} = \dfrac{2t}{1+t^2}$.
Double-angle for $\cos x$:
$\cos x = \cos^2(x/2) - \sin^2(x/2) = \dfrac{1}{1+t^2} - \dfrac{t^2}{1+t^2} = \dfrac{1-t^2}{1+t^2}$.
Differential $dx$: Differentiate $t = \tan(x/2)$ to give $dt = \tfrac{1}{2}\sec^2(x/2)\,dx = \tfrac{1}{2}(1+t^2)\,dx$, so $dx = \dfrac{2}{1+t^2}\,dt$.
$t = \tan(x/2)$ · $\sin x = \dfrac{2t}{1+t^2}$, $\cos x = \dfrac{1-t^2}{1+t^2}$, $dx = \dfrac{2}{1+t^2}\,dt$ · All three derived from the right-triangle and $dt = \tfrac{1}{2}(1+t^2)\,dx$ · Use whenever the integrand is a rational function of $\sin x, \cos x$ and nothing else works
Pause — copy $t = \tan(x/2)$, the three formulas ($\sin x$, $\cos x$, $dx$), and the rule to use $t$-sub when the integrand is a rational function of $\sin x$ and $\cos x$ into your book.
Quick check: Under the substitution $t = \tan(x/2)$, which of the following is the correct expression for $\cos x$?
We just saw the Weierstrass $t$-substitution: $t = \tan(x/2)$ gives $\sin x = 2t/(1+t^2)$, $\cos x = (1-t^2)/(1+t^2)$, $dx = 2\,dt/(1+t^2)$, derived from the right-triangle and $dt/dx = \tfrac{1}{2}(1+t^2)$. That raises a question: once substituted, how do we simplify the resulting rational function in $t$ to a recognisable form? This card answers it → replace all three pieces simultaneously; $1+t^2$ typically cancels in $dx$, leaving $\int A\,dt/(at^2+bt+c)$ → complete the square → arctan.
To evaluate $\displaystyle\int R(\sin x, \cos x)\,dx$ where $R$ is rational, replace each piece:
- $\sin x \to \dfrac{2t}{1+t^2}$
- $\cos x \to \dfrac{1-t^2}{1+t^2}$
- $dx \to \dfrac{2}{1+t^2}\,dt$
Multiply top and bottom of the resulting expression by $1+t^2$ (or some power) to clear all denominators, then integrate the resulting rational function of $t$. Finally back-substitute $t = \tan(x/2)$.
Replace all three pieces — never just $\sin x$ or just $\cos x$ · Multiply through by $1+t^2$ to clear; the $dx$ factor often cancels the $1+t^2$ in the numerator · End-form is usually $\int \dfrac{A\,dt}{at^2 + bt + c}$ → complete the square → $\arctan$ · Back-substitute $t = \tan(x/2)$ at the end (or change limits for definite integrals)
Pause — copy the full substitution procedure (replace all three, clear $1+t^2$, complete the square → arctan), and the back-substitution $t = \tan(x/2)$ at the end into your book.
Did you get this? True or false: when using $t = \tan(x/2)$ on a definite integral from $x = 0$ to $x = \pi/2$, the new limits are $t = 0$ to $t = 1$.
Worked examples · 3 in a row, reveal as you go
Evaluate $\displaystyle\int \frac{1}{2 + \cos x}\,dx$ using $t = \tan(x/2)$.
Evaluate $\displaystyle\int \frac{1}{\sin x + \cos x}\,dx$ using $t = \tan(x/2)$.
Evaluate $\displaystyle\int_0^{\pi/2} \frac{1}{1 + \sin x}\,dx$ using $t = \tan(x/2)$.
Fill the gap: Under $t = \tan(x/2)$: $\sin x = \dfrac{2t}{1+t^2}$ and $dx = \dfrac{}{1+t^2}\,dt$, while $\cos x = \dfrac{1 - }{1+t^2}$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the substitution $t = \tan(x/2)$ is the same as $t = \tan x$ for the purposes of integration.
Activities · practice with the ideas
Apply $t = \tan(x/2)$ to write $\displaystyle\int \frac{1}{5 + 3\cos x}\,dx$ as an integral in $t$. Simplify but do not yet integrate.
Complete the previous integral to find $\displaystyle\int \frac{1}{5+3\cos x}\,dx$.
Evaluate $\displaystyle\int_0^{\pi/2} \frac{1}{1 + \cos x}\,dx$ using $t = \tan(x/2)$ and changing limits.
Use $t = \tan(x/2)$ to evaluate $\displaystyle\int \frac{1}{\sin x}\,dx$ (i.e., $\int \csc x\,dx$) and compare to the standard answer.
Show that $\displaystyle\int_0^{\pi/2} \frac{1}{2 + \sin x}\,dx$ reduces to an arctan form. Set up the rational $t$-integral, then evaluate.
Odd one out: Three of these are correct consequences of $t = \tan(x/2)$. Which one is NOT?
Earlier you wondered what would happen if $\cos x$ could be replaced by a rational function of a new variable.
The Weierstrass substitution $t = \tan(x/2)$ does exactly that: $\cos x$ becomes $\tfrac{1-t^2}{1+t^2}$, $\sin x$ becomes $\tfrac{2t}{1+t^2}$, and $dx$ becomes $\tfrac{2}{1+t^2}\,dt$. The $1+t^2$ factors cancel cleanly between the substituted trig terms and the $dx$ factor, leaving a rational function of $t$ that you can integrate with partial fractions or a standard arctan. Reach for $t = \tan(x/2)$ whenever the integrand mixes $\sin x$ and $\cos x$ in a denominator and no obvious identity helps.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Using $t = \tan(x/2)$, write $\displaystyle\int \frac{1}{3 + \cos x}\,dx$ as an integral in $t$, fully simplified (but do not integrate). (2 marks)
Q2. Evaluate $\displaystyle\int \frac{1}{3 + \cos x}\,dx$ in closed form, expressing the answer in terms of $x$. (3 marks)
Q3. Evaluate $\displaystyle\int_0^{\pi/2} \frac{1}{1 + \sin x + \cos x}\,dx$ using $t = \tan(x/2)$. Show the change of limits explicitly. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $5 + 3\cos x = \dfrac{5(1+t^2) + 3(1-t^2)}{1+t^2} = \dfrac{8 + 2t^2}{1+t^2}$. Integral: $\displaystyle\int \dfrac{1+t^2}{8+2t^2} \cdot \dfrac{2}{1+t^2}\,dt = \int \dfrac{2}{8+2t^2}\,dt = \int \dfrac{dt}{4+t^2}$.
2. $\displaystyle\int \dfrac{dt}{4+t^2} = \dfrac{1}{2}\arctan\!\left(\dfrac{t}{2}\right) + C = \dfrac{1}{2}\arctan\!\left(\dfrac{\tan(x/2)}{2}\right) + C$.
3. Limits $0 \to 1$. $1 + \cos x = \dfrac{2}{1+t^2}$. Integrand: $\dfrac{1+t^2}{2} \cdot \dfrac{2}{1+t^2}\,dt = dt$. So $\displaystyle\int_0^1 dt = 1$.
4. $\displaystyle\int \csc x\,dx = \int \dfrac{(1+t^2)}{2t} \cdot \dfrac{2}{1+t^2}\,dt = \int \dfrac{dt}{t} = \ln|\tan(x/2)| + C$.
5. $2 + \sin x = \dfrac{2t^2 + 2t + 2}{1+t^2}$. Integrand: $\dfrac{2\,dt}{2t^2+2t+2} = \dfrac{dt}{t^2+t+1}$. Complete square: $(t+\tfrac{1}{2})^2 + \tfrac{3}{4}$. $\displaystyle\int_0^1 \dfrac{dt}{(t+1/2)^2 + 3/4} = \dfrac{2}{\sqrt{3}}\left[\arctan\!\left(\dfrac{2t+1}{\sqrt{3}}\right)\right]_0^1 = \dfrac{2}{\sqrt{3}}\left(\arctan\sqrt{3} - \arctan(1/\sqrt{3})\right) = \dfrac{2}{\sqrt{3}}\left(\dfrac{\pi}{3} - \dfrac{\pi}{6}\right) = \dfrac{\pi}{3\sqrt{3}}$.
Q1 (2 marks): Substitute $\cos x$, $dx$, combine over $1+t^2$ [1]: $\displaystyle\int \dfrac{dt}{2 + t^2}$ [1].
Q2 (3 marks): Standard arctan form with $a^2 = 2$ [1]: $\dfrac{1}{\sqrt{2}}\arctan\!\left(\dfrac{t}{\sqrt{2}}\right) + C$ [1]. Back-substitute $t = \tan(x/2)$: $\dfrac{1}{\sqrt{2}}\arctan\!\left(\dfrac{\tan(x/2)}{\sqrt{2}}\right) + C$ [1].
Q3 (3 marks): Limits $x = 0 \to t = 0$; $x = \pi/2 \to t = 1$ [1]. Simplify integrand to $\dfrac{dt}{1+t}$ [1]. $\displaystyle\int_0^1 \dfrac{dt}{1+t} = [\ln(1+t)]_0^1 = \ln 2$ [1].
Five timed questions on the t-substitution — setup, simplification, and arctan/log forms. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering quick t-substitution questions. Lighter alternative to the boss.
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