Conquering the Impossible Integral: A Guide to the t-Substitution

If you are navigating the HSC Mathematics Extension 2 syllabus, you already know that integration is as much an art as it is a science. While standard techniques like integration by parts, partial fractions, and reverse chain rule can solve a wide array of problems, there is a specific class of trigonometric integrals that stubbornly resists these standard tools.

Enter the $t$-substitution (often known in university calculus as the Weierstrass substitution). It is the ultimate heavy-duty tool for dealing with messy rational trigonometric expressions. Let’s break down how it works and walk through some classic examples.

What is it, and why is it useful?

At its core, the $t$-substitution is a clever algebraic bridge. It transforms a complex trigonometric integrand (involving $\sin x$ and $\cos x$ in the denominators) into a standard rational function (a polynomial divided by a polynomial).

Why is this a lifesaver? Because rational functions can almost always be integrated using algebraic techniques you already know, such as partial fractions, completing the square, or recognizing standard inverse tangent and logarithmic forms. Once the substitution is made, the trigonometry disappears entirely until the final step.

Quick Review of the Substitution

As outlined in the reference guide, the substitution relies on setting $t = \tan\left(\frac{x}{2}\right)$. Using double-angle formulas, we can express $\sin x$, $\cos x$, and $\tan x$ strictly in terms of $t$.

Here is your essential toolkit to memorize: Let $t = \tan\left(\frac{x}{2}\right)$. Then:

• $\sin x = \frac{2t}{1+t^2}$
• $\cos x = \frac{1-t^2}{1+t^2}$
• $\tan x = \frac{2t}{1-t^2}$
• $dx = \frac{2}{1+t^2} \, dt$

Note: Don't forget to swap out the $dx$! The differential $dx$ carries its own denominator of $1+t^2$, which usually elegantly cancels out the denominators introduced by the $\sin x$ and $\cos x$ substitutions.

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Solving the Problems

Let’s apply this to a customized practice set based on the structures in image_8a0f0a.png. We will tackle a selection of both indefinite and definite integrals to see the mechanics in action.

Part 1: Indefinite Integrals

Problem 1a (Modified): Find $\int \frac{1}{2+\cos x} \, dx$

1. Substitute: Replace $\cos x$ and $dx$ with their $t$-equivalents.

$$\int \frac{1}{2 + \left(\frac{1-t^2}{1+t^2}\right)} \cdot \frac{2}{1+t^2} \, dt$$

2. Simplify: Multiply the top and bottom of the first fraction by $1+t^2$.

$$\int \frac{2}{2(1+t^2) + (1-t^2)} \, dt = \int \frac{2}{2 + 2t^2 + 1 - t^2} \, dt = \int \frac{2}{t^2 + 3} \, dt$$

3. Integrate: This fits the standard inverse tangent form $\int \frac{1}{t^2 + a^2} dt = \frac{1}{a}\arctan\left(\frac{t}{a}\right)$, where $a = \sqrt{3}$.

$$\frac{2}{\sqrt{3}} \arctan\left(\frac{t}{\sqrt{3}}\right) + C$$

4. Back-substitute: Swap $t$ back for $x$.

$$\frac{2}{\sqrt{3}} \arctan\left(\frac{\tan(x/2)}{\sqrt{3}}\right) + C$$

Problem 1b (Modified): Find $\int \frac{1}{\frac{1}{2}+\sin x - \cos x} \, dx$

1. Substitute:

$$\int \frac{1}{\frac{1}{2} + \frac{2t}{1+t^2} - \frac{1-t^2}{1+t^2}} \cdot \frac{2}{1+t^2} \, dt$$

2. Simplify: Multiply the numerator and denominator by $1+t^2$.

$$\int \frac{2}{\frac{1}{2}(1+t^2) + 2t - (1-t^2)} \, dt = \int \frac{2}{\frac{1}{2} + \frac{1}{2}t^2 + 2t - 1 + t^2} \, dt$$ $$= \int \frac{2}{\frac{3}{2}t^2 + 2t - \frac{1}{2}} \, dt$$

Multiply the entire fraction by $\frac{2}{2}$ to clear the denominators:

$$\int \frac{4}{3t^2 + 4t - 1} \, dt$$

3. Complete the Square: Let's prepare the denominator for an algebraic integration technique.

$$3t^2 + 4t - 1 = 3\left(t^2 + \frac{4}{3}t\right) - 1 = 3\left[\left(t + \frac{2}{3}\right)^2 - \frac{4}{9}\right] - 1 = 3\left(t + \frac{2}{3}\right)^2 - \frac{7}{3}$$

Pulling out the common factor of $3$:

$$\frac{4}{3} \int \frac{1}{(t + 2/3)^2 - (\sqrt{7}/3)^2} \, dt$$

4. Integrate: Using the standard standard integral form $\int \frac{1}{u^2 - a^2} du = \frac{1}{2a}\ln\left\vert{}\frac{u-a}{u+a}\right\vert{}$. Here, $u = t + \frac{2}{3}$ and $a = \frac{\sqrt{7}}{3}$.

$$= \frac{4}{3} \cdot \frac{1}{2(\sqrt{7}/3)} \ln\left\vert{} \frac{t + 2/3 - \sqrt{7}/3}{t + 2/3 + \sqrt{7}/3} \right\vert{} + C$$ $$= \frac{2}{\sqrt{7}} \ln\left\vert{} \frac{3t + 2 - \sqrt{7}}{3t + 2 + \sqrt{7}} \right\vert{} + C$$

5. Back-substitute:

$$\frac{2}{\sqrt{7}} \ln\left\vert{} \frac{3\tan(x/2) + 2 - \sqrt{7}}{3\tan(x/2) + 2 + \sqrt{7}} \right\vert{} + C$$

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Part 2: Definite Integrals

Crucial Step: For definite integrals, you MUST change your limits of integration from $x$-values to $t$-values. This saves you from having to back-substitute at the end.

Problem 2a (Modified): Evaluate $\int_0^{\pi/2} \frac{1}{1-\sin x} \, dx$

1. Change the limits:

• When $x = 0$, $t = \tan(0/2) = 0$.
• When $x = \pi/2$, $t = \tan(\pi/4) = 1$.

2. Substitute:

$$\int_0^1 \frac{1}{1 - \frac{2t}{1+t^2}} \cdot \frac{2}{1+t^2} \, dt$$

3. Simplify: Multiply through by $1+t^2$.

$$\int_0^1 \frac{2}{(1+t^2) - 2t} \, dt = \int_0^1 \frac{2}{(t-1)^2} \, dt$$

4. Evaluate:

$$\int_0^1 2(t-1)^{-2} \, dt = \left[ -2(t-1)^{-1} \right]_0^1 = \left[ \frac{-2}{t-1} \right]_0^1$$

Note: At $t=1$, this expression approaches infinity. Looking closely at our original integrand, as $x \to \pi/2$, $\sin x \to 1$, making the denominator $0$. This is an improper integral!

$$\lim_{k \to 1^-} \left[ \frac{-2}{t-1} \right]_0^k = \lim_{k \to 1^-} \left( \frac{-2}{k-1} \right) - \left( \frac{-2}{0-1} \right) = \infty - 2 = \infty$$

Takeaway: The integral diverges! Recognising domain limitations and asymptote behaviors under substitutions is a hallmark skill for Extension 2.

Problem 2c (Modified): Evaluate $\int_{0}^{\pi/2} \frac{1}{3+\sin x} \, dx$

1. Change the limits:

• When $x = 0$, $t = \tan(0) = 0$.
• When $x = \pi/2$, $t = \tan(\pi/4) = 1$.

2. Substitute & Simplify:

$$\int_{0}^1 \frac{1}{3 + \left(\frac{2t}{1+t^2}\right)} \cdot \frac{2}{1+t^2} \, dt = \int_{0}^1 \frac{2}{3(1+t^2) + 2t} \, dt$$ $$= \int_{0}^1 \frac{2}{3t^2 + 2t + 3} \, dt$$

3. Complete the Square: Factor out the 3: $3\left(t^2 + \frac{2}{3}t + 1\right)$ Complete the square: $3\left[ \left(t + \frac{1}{3}\right)^2 - \frac{1}{9} + \frac{9}{9} \right] = 3\left[ \left(t + \frac{1}{3}\right)^2 + \frac{8}{9} \right]$
4. Integrate (Standard Inverse Tangent Form):

$$\frac{2}{3} \int_{0}^1 \frac{1}{(t + 1/3)^2 + (\sqrt{8}/3)^2} \, dt$$

Using $\int \frac{1}{u^2 + a^2} du = \frac{1}{a} \arctan\left(\frac{u}{a}\right)$, where $a = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$.

$$= \frac{2}{3} \cdot \frac{3}{2\sqrt{2}} \left[ \arctan\left(\frac{t + 1/3}{2\sqrt{2}/3}\right) \right]_{0}^1 = \frac{1}{\sqrt{2}} \left[ \arctan\left(\frac{3t + 1}{2\sqrt{2}}\right) \right]_{0}^1$$

5. Evaluate:

$$= \frac{1}{\sqrt{2}} \left( \arctan\left(\frac{4}{2\sqrt{2}}\right) - \arctan\left(\frac{1}{2\sqrt{2}}\right) \right)$$ $$= \frac{1}{\sqrt{2}} \left( \arctan(\sqrt{2}) - \arctan\left(\frac{1}{2\sqrt{2}}\right) \right)$$

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Key Takeaways for the Exam

When utilizing the $t$-substitution in an Extension 2 environment, keep this checklist in mind:

• Identify the Trigger: Look for integrands with $a\sin x + b\cos x + c$ in the denominator.
• Don't Forget $dx$: The most common student error is replacing $\sin x$ and $\cos x$ but leaving $dx$ as is. Always include the $\frac{2}{1+t^2}dt$ multiplier!
• Definite Integrals = New Limits: Always update your limits of integration immediately. It saves time and drastically reduces algebraic errors at the end.
• Watch the Boundaries: Pay close attention to integration limits like $\pi$ or values where the denominator becomes $0$, as they can lead to undefined scenarios or improper integrals.

Conclusion

The $t$-substitution can seem intimidating at first glance because it involves memorizing formulas and handling dense algebra. However, once you master the setup, it reduces complex trigonometry problems into predictable rational functions. Practice setting up the algebraic steps clearly—your marker will appreciate a well-structured response just as much as a correct final answer. Happy integrating!