Ever stared at a trig‑heavy integral and thought, “There’s got to be a simpler way?”
You’re not alone. The moment you replace the sine or cosine with a single variable — usually called u — the whole problem can collapse into plain algebra. It feels like magic, but it’s just a tidy change of perspective.
Below is everything you need to turn those tangled trigonometric expressions into clean algebraic ones, step by step. No fluff, just the real‑talk shortcuts that actually work.
What Is “Write the Trigonometric Expression as an Algebraic Expression in u”?
In practice, this phrase means you’re taking a formula that involves * sin θ*, * cos θ*, * tan θ* (or any combination) and substituting a new variable, u, for one of those trig functions. Think about it: the goal? Turn the whole thing into a polynomial or rational expression in u that you can manipulate with ordinary algebra That's the part that actually makes a difference..
Typical choices for u are:
- u = sin θ – great when the expression also contains cos θ or √(1‑sin²θ).
- u = cos θ – handy when you have sin θ or tan θ.
- u = tan(θ/2) – the classic Weierstrass substitution; it converts every trig function into a rational function of u.
Once you pick the right substitution, you rewrite every trig piece in terms of u, simplify, and you’ve got an algebraic expression that’s ready for factoring, completing the square, or whatever the problem calls for And that's really what it comes down to..
Why It Matters / Why People Care
Because trig can feel like a maze. One minute you’re staring at sec θ + tan θ, the next you need an antiderivative for a calculus exam. If you can rewrite the whole thing as a polynomial in u, the maze disappears That's the whole idea..
- Integration – many integrals that look impossible become straightforward rational integrals after a u‑substitution.
- Equation solving – turning sin θ = ½ into u = ½ lets you use quadratic formulas instead of guessing angles.
- Simplification – in physics or engineering, you often need a compact form for a transfer function. Rewriting in u cuts down on parentheses and makes error‑checking easier.
Bottom line: you trade a trigonometric headache for an algebraic one, and algebra is usually the easier side to be on.
How It Works (or How to Do It)
Below are the three most common routes. Pick the one that matches your problem, follow the steps, and you’ll end up with a tidy algebraic expression It's one of those things that adds up..
1. Simple Substitution (u = sin θ or u = cos θ)
Step‑by‑step
- Identify the dominant trig function – Is the expression dominated by sin θ or cos θ?
- Set u – Let u = sin θ or u = cos θ.
- Express the companion function – Use the Pythagorean identity sin²θ + cos²θ = 1.
- If u = sin θ, then cos θ = ±√(1 – u²).
- If u = cos θ, then sin θ = ±√(1 – u²).
- Replace every trig term – Plug u and the square‑root expression wherever the original functions appear.
- Simplify – Rationalize the denominator if needed, combine like terms, and you’ve got a pure algebraic expression.
Example
Write ( \displaystyle \frac{\sin\theta}{1+\cos\theta} ) as an algebraic expression in u with u = \sin\theta.
Step 1: u = sin θ → cos θ = √(1 – u²) (choose the sign based on the quadrant; assume positive for now).
Step 2: Substitute:
[ \frac{u}{1+\sqrt{1-u^{2}}} ]
Step 3: Multiply numerator and denominator by the conjugate to eliminate the root:
[ \frac{u\bigl(1-\sqrt{1-u^{2}}\bigr)}{(1+\sqrt{1-u^{2}})(1-\sqrt{1-u^{2}})} = \frac{u\bigl(1-\sqrt{1-u^{2}}\bigr)}{1-(1-u^{2})} = \frac{u\bigl(1-\sqrt{1-u^{2}}\bigr)}{u^{2}} ]
Cancel one u:
[ \frac{1-\sqrt{1-u^{2}}}{u} ]
That’s the algebraic form. You can stop here or rationalize again, depending on what you need Worth keeping that in mind. Still holds up..
2. The Weierstrass Substitution (u = tan (θ/2))
This is the heavy‑duty tool that turns any trig expression into a rational function of u. It’s a bit longer, but once you memorize the three core identities, you’ll be unstoppable It's one of those things that adds up. Practical, not theoretical..
Core identities
[ \sin\theta = \frac{2u}{1+u^{2}},\qquad \cos\theta = \frac{1-u^{2}}{1+u^{2}},\qquad \tan\theta = \frac{2u}{1-u^{2}} ]
where u = \tan(\theta/2).
Step‑by‑step
- Set u = tan(θ/2).
- Replace each trig function using the identities above.
- Simplify the rational expression – combine fractions, cancel common factors, and you’ll have a polynomial or proper rational function in u.
- (Optional) Back‑substitute if you need the answer in terms of θ later.
Example
Convert ( \displaystyle \frac{1-\cos\theta}{\sin\theta} ) into an algebraic expression in u.
Step 1: Use the identities:
[ 1-\cos\theta = 1 - \frac{1-u^{2}}{1+u^{2}} = \frac{(1+u^{2})-(1-u^{2})}{1+u^{2}} = \frac{2u^{2}}{1+u^{2}} ]
[ \sin\theta = \frac{2u}{1+u^{2}} ]
Step 2: Form the quotient:
[ \frac{2u^{2}/(1+u^{2})}{2u/(1+u^{2})} = \frac{2u^{2}}{1+u^{2}} \cdot \frac{1+u^{2}}{2u} = \frac{u^{2}}{u} = u ]
So the whole expression collapses to simply u. That’s the power of the Weierstrass trick.
3. Mixed Substitution (u = tan θ, u = sec θ, etc.)
Sometimes the problem naturally suggests a different substitution, especially when the expression already contains a tangent or secant.
Quick guide
- u = tan θ – use ( \sec^{2}\theta = 1 + \tan^{2}\theta = 1 + u^{2} ).
- u = sec θ – use ( \tan^{2}\theta = \sec^{2}\theta - 1 = u^{2} - 1 ).
Replace the chosen function and rewrite the rest with the Pythagorean identities.
Example
Write ( \displaystyle \frac{\tan\theta}{\sec\theta + 1} ) as an algebraic expression in u with u = \tan\theta.
Step 1: ( \sec\theta = \sqrt{1+u^{2}} ) (again, sign depends on quadrant).
Step 2: Substitute:
[ \frac{u}{\sqrt{1+u^{2}} + 1} ]
Step 3: Rationalize the denominator:
[ \frac{u\bigl(\sqrt{1+u^{2}} - 1\bigr)}{( \sqrt{1+u^{2}}+1)(\sqrt{1+u^{2}}-1)} = \frac{u\bigl(\sqrt{1+u^{2}} - 1\bigr)}{(1+u^{2}) - 1} = \frac{u\bigl(\sqrt{1+u^{2}} - 1\bigr)}{u^{2}} ]
Cancel one u:
[ \frac{\sqrt{1+u^{2}} - 1}{u} ]
That’s the algebraic version. It’s ready for further manipulation—maybe a series expansion, maybe a limit.
Common Mistakes / What Most People Get Wrong
- Forgetting the ± sign – When you replace cos θ with √(1‑u²), the sign isn’t always positive. Ignoring the quadrant leads to wrong answers, especially in definite integrals.
- Mixing up the Weierstrass identities – It’s easy to swap the numerator and denominator for tan θ. Double‑check: tan θ = 2u/(1‑u²), not the other way around.
- Leaving hidden roots – After substitution, you often get √(1‑u²). If you can rationalize or square both sides early, you avoid messy radicals later.
- Skipping the Jacobian – In integration, the differential dθ must be transformed: dθ = 2 du/(1+u²) when u = tan(θ/2). Forgetting this factor throws the whole integral off.
- Assuming u is small – Some textbooks treat u as a “tiny” quantity to linearize, but the substitution is exact. Don’t truncate unless you’re explicitly doing a series approximation.
Practical Tips / What Actually Works
- Pick the substitution that eliminates the most terms. If the expression already has a tan θ, go with u = tan θ instead of u = sin θ.
- Write down the identity sheet before you start. A quick reference to the three Weierstrass formulas saves you from hunting through notes mid‑problem.
- Check the domain early. Knowing whether θ lies in [0, π/2] or another interval tells you which sign to use for the square root.
- When rationalizing, aim for the conjugate—it’s the fastest way to clear a root in the denominator.
- If you end up with a quadratic in u, solve it with the standard formula; you’ll often get two possible u values, which correspond to two angles.
- Test with a numeric angle. Plug θ = 30° into both the original and the algebraic form; they should match. It’s a quick sanity check before you move on.
- Keep the original trig expression handy until you’re sure the algebraic version is correct. It’s easy to lose a sign or a factor in the shuffle.
FAQ
Q1: When should I use the Weierstrass substitution versus a simple u = sin θ?
A: If the expression contains a mix of sin, cos, and tan or if you need to integrate a rational function of trig, the Weierstrass substitution is the cleanest. For a single‑function problem, a direct u = sin θ or u = cos θ is quicker.
Q2: Does the substitution work for inverse trig functions?
A: Yes, but you typically substitute after you’ve expressed the inverse trig in terms of a regular trig function. Take this: arcsin x becomes θ with sin θ = x, then you can set u = x directly Nothing fancy..
Q3: How do I handle definite integrals after substitution?
A: Convert the limits: if θ = a and θ = b, compute u = tan(a/2) and u = tan(b/2). Then integrate with respect to u using the transformed limits That alone is useful..
Q4: What if the square root √(1‑u²) stays in the denominator?
A: Multiply numerator and denominator by the conjugate (√(1‑u²) − 1) to eliminate the root, or square both sides if you’re solving an equation.
Q5: Is there a “one‑size‑fits‑all” substitution?
A: Not really. The best choice depends on which trig functions dominate the expression. The trick is to recognize patterns quickly—once you do, the right substitution pops out automatically Practical, not theoretical..
Turning trigonometric clutter into neat algebra isn’t a magic trick; it’s a systematic rewrite. Pick the right u, apply the corresponding identities, watch the radicals disappear, and you’ll be solving integrals or equations in minutes instead of hours Easy to understand, harder to ignore..
Give it a try on your next calculus problem—you’ll see how satisfying it is to watch a messy sine‑cosine mess melt into a simple polynomial. Happy substituting!
A Worked‑Out Example (Putting It All Together)
Suppose you’re asked to evaluate
[ I=\int_{0}^{\pi/4}\frac{1+\sin\theta}{\cos\theta+\sqrt{1-\sin^{2}\theta}};d\theta . ]
At first glance the denominator looks intimidating because of the square‑root term, but notice that
[ \sqrt{1-\sin^{2}\theta}=|\cos\theta| . ]
Since the interval ([0,\pi/4]) lies entirely in the first quadrant, (\cos\theta\ge 0) and the absolute value can be dropped. The integral simplifies to
[ I=\int_{0}^{\pi/4}\frac{1+\sin\theta}{\cos\theta+\cos\theta};d\theta =\int_{0}^{\pi/4}\frac{1+\sin\theta}{2\cos\theta};d\theta . ]
Now we have a rational combination of (\sin\theta) and (\cos\theta). The classic choice is the Weierstrass substitution
[ u=\tan\frac{\theta}{2},\qquad \sin\theta=\frac{2u}{1+u^{2}},\qquad \cos\theta=\frac{1-u^{2}}{1+u^{2}},\qquad d\theta=\frac{2,du}{1+u^{2}} . ]
Step 1 – Change the limits.
[ \theta=0;\Rightarrow;u=\tan0=0,\qquad \theta=\frac{\pi}{4};\Rightarrow;u=\tan\frac{\pi}{8}= \sqrt{2}-1 . ]
Step 2 – Substitute.
[ \begin{aligned} I &=\int_{0}^{\sqrt{2}-1} \frac{1+\displaystyle\frac{2u}{1+u^{2}}} {2\displaystyle\frac{1-u^{2}}{1+u^{2}}} \cdot\frac{2,du}{1+u^{2}}\[4pt] &=\int_{0}^{\sqrt{2}-1} \frac{(1+u^{2})+2u}{2(1-u^{2})}, \frac{2,du}{(1+u^{2})^{2}}\[4pt] &=\int_{0}^{\sqrt{2}-1} \frac{1+u^{2}+2u}{(1-u^{2})(1+u^{2})^{2}},du . \end{aligned} ]
Step 3 – Partial‑fraction decomposition.
Write
[ \frac{1+u^{2}+2u}{(1-u^{2})(1+u^{2})^{2}} =\frac{A}{1-u}+ \frac{B}{1+u} +\frac{Cu+D}{1+u^{2}} . ]
Solving for the constants (a quick algebraic grind) yields
[ A=\frac14,\qquad B=-\frac14,\qquad C=0,\qquad D=\frac12 . ]
Thus
[ I=\int_{0}^{\sqrt{2}-1} \Bigl[\frac{1/4}{1-u} -\frac{1/4}{1+u} +\frac{1/2}{1+u^{2}}\Bigr]du . ]
Step 4 – Integrate term by term.
[ \begin{aligned} I &=\frac14\bigl[-\ln|1-u|-\ln|1+u|\bigr]{0}^{\sqrt{2}-1} +\frac12\bigl[\arctan u\bigr]{0}^{\sqrt{2}-1}\[4pt] &=\frac14\Bigl[\ln\frac{1+u}{1-u}\Bigr]_{0}^{\sqrt{2}-1} +\frac12\arctan(\sqrt{2}-1) . \end{aligned} ]
Plugging the limits in:
[ \begin{aligned} \ln\frac{1+(\sqrt{2}-1)}{1-(\sqrt{2}-1)} &=\ln\frac{\sqrt{2}}{2-\sqrt{2}} =\ln\bigl(\sqrt{2}+1\bigr) ,\[4pt] \arctan(\sqrt{2}-1)&=\frac{\pi}{8}. \end{aligned} ]
Finally,
[ \boxed{I=\frac14\ln\bigl(\sqrt{2}+1\bigr)+\frac{\pi}{16}} . ]
The whole computation hinged on three small decisions:
- Recognize the hidden absolute value and simplify the radical early.
- Choose the Weierstrass substitution because the integrand became a rational function of (\sin\theta) and (\cos\theta).
- Convert the limits before integrating, which avoids the extra step of back‑substituting (u) at the end.
TL;DR Checklist for Future Problems
| Situation | Preferred Substitution | Quick Reason |
|---|---|---|
| Only (\sin\theta) or (\cos\theta) appears linearly | (u=\sin\theta) or (u=\cos\theta) | Direct algebraic conversion |
| Mix of (\sin\theta,\cos\theta,\tan\theta) in a rational expression | (u=\tan\frac{\theta}{2}) (Weierstrass) | Turns every trig function into a rational function of (u) |
| Presence of (\sqrt{1-\sin^{2}\theta}) or (\sqrt{1-\cos^{2}\theta}) | First test sign of the underlying cosine/sine; replace with ( | \cos\theta |
| Definite integral with limits that are “nice” multiples of (\pi) | Any substitution, but always transform limits immediately | Keeps the final answer in the same angular domain and avoids back‑substitution errors |
| Quadratic in the new variable after substitution | Solve with the quadratic formula; keep both roots until the original domain discards one | Guarantees you don’t lose a legitimate solution prematurely |
Closing Thoughts
The art of trigonometric substitution is less about memorising a handful of formulas and more about cultivating a pattern‑recognition mindset. When you see a square root of a sine‑squared or cosine‑squared term, ask yourself whether the absolute‑value interpretation clears the radical. When you encounter a blend of sine, cosine, and tangent, reach for the Weierstrass substitution and let the algebra do the heavy lifting.
With the quick‑reference tips, the FAQ clarifications, and the step‑by‑step example above, you now have a compact toolbox that fits in the back pocket of any calculus exam or homework session. The next time a trig‑laden integral or equation blocks your progress, pause, pick the appropriate u, rewrite, and watch the problem dissolve into a familiar polynomial form.
Happy solving—and may your substitutions always be smooth!
5. When the “obvious” substitution fails – a fallback plan
Even after a careful scan, you might find that none of the three standard tricks (simple (u=\sin\theta) or (u=\cos\theta), the Weierstrass change, or an absolute‑value simplification) yields a tractable integral. In that case, keep these two safety nets in mind:
-
Partial‑fraction decomposition after a rationalizing multiply
If the integrand contains (\sqrt{a\pm b\sin\theta}) or (\sqrt{a\pm b\cos\theta}) with constants (a,b), try multiplying numerator and denominator by the conjugate. The resulting expression often becomes a sum of a rational function and a simpler radical that can be handled by the first two substitutions. -
Series expansion for “small‑angle’’ limits
When the limits are close to a point where the integrand behaves nicely (e.g., (\theta\to0) or (\theta\to\pi/2)), expand the trigonometric functions to the first few terms. The integral of the truncated series can be evaluated analytically, and the error can be bounded using the remainder term of the Taylor series. This is especially handy in physics problems where an exact closed form is unnecessary Nothing fancy..
Both strategies are “last‑resort” because they either introduce extra algebraic baggage (the conjugate) or sacrifice exactness (the series). That said, they give you a systematic route forward when the direct path is blocked And that's really what it comes down to..
6. A compact “cheat‑sheet” you can paste on a notebook
Below is a one‑page summary that captures everything we have discussed. Feel free to copy it into a margin or a sticky note.
TRIG SUBSTITUTION QUICK REFERENCE
1. Identify the radical:
√(1‑sin²θ) → |cosθ|
√(1‑cos²θ) → |sinθ|
√(a±b sinθ) or √(a±b cosθ) → try conjugate
2. Choose substitution:
• u = sinθ (if only sin appears linearly)
• u = cosθ (if only cos appears linearly)
• u = tan(θ/2) (Weierstrass) → sinθ = 2u/(1+u²), cosθ = (1‑u²)/(1+u²),
dθ = 2 du/(1+u²)
3. Convert limits immediately:
θ = α → u = tan(α/2) (or u = sinα, u = cosα as appropriate)
4. Simplify the integrand:
• Replace |cosθ|, |sinθ| by the appropriate sign on the interval.
• Rationalize any remaining radicals.
5. Integrate:
• For rational functions → partial fractions.
• For logarithmic forms → ∫ du/(u±c) = ln|u±c|.
• For arctan forms → ∫ du/(1+u²) = arctan u.
6. Back‑substitute (if needed) and evaluate at the new limits.
7. Verify:
• Check sign of the original radical on the interval.
• Differentiate the result to recover the integrand (optional).
7. Final remarks
The journey from the original integral
[ I=\int_{0}^{\pi/4}\frac{\sqrt{1-\sin^{2}\theta}}{1+\sin\theta},d\theta ]
to the tidy closed‑form
[ \boxed{I=\frac14\ln\bigl(\sqrt{2}+1\bigr)+\frac{\pi}{16}} ]
illustrates a universal pattern:
- Spot the hidden simplification (the absolute value of a cosine).
- Pick the substitution that linearises the remaining trigonometric part (the Weierstrass change).
- Translate the limits early to stay in the algebraic world.
When you internalise these three moves, most trigonometric integrals that look intimidating at first glance will collapse into elementary logarithms and arctangents. The checklist, FAQ, and cheat‑sheet above are designed to make that mental workflow automatic And it works..
So the next time a problem asks you to “evaluate the integral involving a square root of a sine‑squared term,” remember: simplify the radical, substitute wisely, and never forget to move the bounds. With those habits, the integral will almost solve itself.
Happy integrating!
7. A “what‑if” toolbox – extending the method to related integrals
The techniques above are not limited to the specific combination of (\sqrt{1-\sin^{2}\theta}) and (1+\sin\theta). In practice you will often encounter variants that differ only by a constant factor, a sign change, or a shifted interval. Below are a few common “what‑if” scenarios and the minimal tweak required to reuse the cheat‑sheet verbatim.
| Variant of the original integral | Immediate observation | Modified step |
|---|---|---|
| (\displaystyle\int_{0}^{\pi/4}\frac{\sqrt{1-\cos^{2}\theta}}{1+\cos\theta},d\theta) | (\sqrt{1-\cos^{2}\theta}= | \sin\theta |
| (\displaystyle\int_{0}^{\pi/4}\frac{\sqrt{1-k^{2}\sin^{2}\theta}}{1+\sin\theta},d\theta) with (0<k<1) | A parameter appears under the radical. On top of that, | |
| (\displaystyle\int_{a}^{b}\frac{\sqrt{1-\sin^{2}\theta}}{c+\sin\theta},d\theta) | The denominator is a linear shift of (\sin\theta). Which means | Split the integral at (\theta=\pi/2) (or at the zero of (\cos\theta)) and treat each subinterval with the appropriate sign for ( |
| (\displaystyle\int_{0}^{\pi/2}\frac{\sqrt{1-\sin^{2}\theta}}{1+\sin\theta},d\theta) | The interval now contains the point where (\cos\theta) changes sign. | |
| (\displaystyle\int_{0}^{\pi/4}\frac{\sqrt{1-\sin^{2}\theta}}{1+\sin^{2}\theta},d\theta) | The denominator now involves (\sin^{2}\theta). | First rationalise the radical by multiplying numerator and denominator by the conjugate (\sqrt{1-k^{2}\sin^{2}\theta}-\sin\theta). |
The key message is that once the radical has been tamed, the Weierstrass substitution turns any expression built from (\sin\theta) and (\cos\theta) into a rational function of (u). After that point the toolbox of elementary integration (partial fractions, standard logarithmic and arctangent antiderivatives) is sufficient Worth keeping that in mind..
8. Common pitfalls and how to avoid them
| Pitfall | Why it hurts | Remedy |
|---|---|---|
| Forgetting the absolute value when (\cos\theta) crosses zero. | You may end up with a sign error that propagates through the whole calculation. Now, | Always check the sign of the underlying trigonometric function on each subinterval before dropping the absolute value. Worth adding: |
| Mixing up the Weierstrass formulas for (\sin\theta) and (\cos\theta). | A swapped numerator/denominator leads to an algebraic mess that no longer simplifies. | Keep a small reference card: (\sin\theta=\frac{2u}{1+u^{2}},;\cos\theta=\frac{1-u^{2}}{1+u^{2}},;d\theta=\frac{2,du}{1+u^{2}}). |
| Not converting the limits before simplifying the integrand. | The sign of (\cos\theta) (or (\sin\theta)) may change after you have already assumed a particular sign, invalidating the simplification. Still, | Perform the substitution including the limits immediately after Step 2; only then proceed to algebraic reduction. |
| Over‑rationalising the integrand by multiplying by unnecessary conjugates. | Extra factors can introduce spurious poles that complicate the partial‑fraction step. On top of that, | Apply the conjugate only when it eliminates a square root that cannot be removed otherwise. |
| Ignoring the domain of the substitution (e.Consider this: g. , (u=\tan(\theta/2)) is undefined at (\theta=\pi)). And | You may inadvertently introduce a discontinuity that the original integral never had. | Verify that the interval you are integrating over lies entirely within a single branch of the substitution; if not, split the integral. |
9. Putting it all together – a worked‑out “quick‑fire” example
Let us illustrate the whole pipeline with a fresh integral that is deliberately similar but not identical to the one that motivated the article:
[ J=\int_{\pi/6}^{\pi/3}\frac{\sqrt{1-\cos^{2}\theta}}{2-\sin\theta},d\theta . ]
Step 1 – Simplify the radical.
(\sqrt{1-\cos^{2}\theta}=|\sin\theta|). On ([\pi/6,\pi/3]) we have (\sin\theta>0), so (|\sin\theta|=\sin\theta).
Step 2 – Choose the substitution.
Because the remaining denominator contains (\sin\theta) linearly, we set (u=\tan(\theta/2)). Then
[ \sin\theta=\frac{2u}{1+u^{2}},\qquad d\theta=\frac{2,du}{1+u^{2}}. ]
Step 3 – Convert limits.
[ \theta=\frac{\pi}{6};\Longrightarrow;u_{1}=\tan!\Bigl(\frac{\pi}{12}\Bigr), \qquad \theta=\frac{\pi}{3};\Longrightarrow;u_{2}=\tan!\Bigl(\frac{\pi}{6}\Bigr)=\frac{1}{\sqrt{3}}. ]
Both are positive, so no sign issues arise.
Step 4 – Rewrite the integrand.
[ J=\int_{u_{1}}^{u_{2}} \frac{\displaystyle\frac{2u}{1+u^{2}}} {,2-\displaystyle\frac{2u}{1+u^{2}}} \cdot\frac{2,du}{1+u^{2}}
\int_{u_{1}}^{u_{2}} \frac{4u}{(1+u^{2})\bigl[,2(1+u^{2})-2u,\bigr]},du
\int_{u_{1}}^{u_{2}} \frac{2u}{(1+u^{2})(1+u^{2}-u)},du . ]
Step 5 – Partial‑fraction decomposition.
Write
[ \frac{2u}{(1+u^{2})(1+u^{2}-u)}
\frac{A u + B}{1+u^{2}}+\frac{C u + D}{1+u^{2}-u}. ]
Solving for (A,B,C,D) yields (A=1,;B=0,;C=-1,;D=1). Hence
[ \frac{2u}{(1+u^{2})(1+u^{2}-u)}
\frac{u}{1+u^{2}}-\frac{u-1}{1+u^{2}-u}. ]
Step 6 – Integrate term by term.
[ \int\frac{u}{1+u^{2}},du=\frac12\ln(1+u^{2}), \qquad \int\frac{u-1}{1+u^{2}-u},du = \frac12\ln!\bigl(1+u^{2}-u\bigr). ]
Thus
[ J=\Bigl[\tfrac12\ln(1+u^{2})-\tfrac12\ln!\bigl(1+u^{2}-u\bigr)\Bigr]{u{1}}^{u_{2}}
\frac12\ln!\frac{1+u^{2}}{1+u^{2}-u}\Biggr|{u{1}}^{u_{2}}. ]
Step 7 – Insert the limits.
[ u_{2}= \frac{1}{\sqrt3}; \Rightarrow; \frac{1+u_{2}^{2}}{1+u_{2}^{2}-u_{2}} = \frac{1+1/3}{1+1/3-1/\sqrt3} = \frac{4/3}{4/3-1/\sqrt3} = \frac{4}{4-\sqrt3}. ]
[ u_{1}= \tan(\pi/12)=2-\sqrt3; \Rightarrow; \frac{1+u_{1}^{2}}{1+u_{1}^{2}-u_{1}} = \frac{1+(2-\sqrt3)^{2}}{1+(2-\sqrt3)^{2}-(2-\sqrt3)} = \frac{1+7-4\sqrt3}{1+7-4\sqrt3-(2-\sqrt3)} = \frac{8-4\sqrt3}{6-3\sqrt3} = \frac{4(2-\sqrt3)}{3(2-\sqrt3)}=\frac{4}{3}. ]
Finally
[ J=\frac12\Bigl[\ln!\Bigl(\frac{4}{4-\sqrt3}\Bigr)-\ln!\Bigl(\frac{4}{3}\Bigr)\Bigr] = \frac12\ln!\Bigl(\frac{3}{4-\sqrt3}\Bigr) = \frac12\ln!\Bigl(\frac{3(4+\sqrt3)}{(4-\sqrt3)(4+\sqrt3)}\Bigr) = \frac12\ln!\Bigl(\frac{12+3\sqrt3}{7}\Bigr). ]
So the integral evaluates to
[ \boxed{J=\tfrac12\ln!\bigl(\tfrac{12+3\sqrt3}{7}\bigr)}. ]
The whole computation mirrors the workflow laid out in the cheat‑sheet; the only extra step was a brief algebraic simplification of the limit expressions. This example demonstrates that once the radical is eliminated, the remaining algebraic work is routine.
10. Conclusion
Trigonometric integrals that hide a square‑root of a sine or cosine often look forbidding at first glance, but they are rarely mysterious. The “secret sauce” consists of three disciplined moves:
- Turn the radical into an absolute value and decide its sign on the interval.
- Apply the universal Weierstrass substitution (or a simpler linear substitution when the structure permits) before you start expanding.
- Convert the limits together with the variable so that you never have to revert to the original angle.
When these steps are followed, the integrand collapses to a rational function of the new variable, and the integration finishes with elementary logarithms and arctangents. The compact cheat‑sheet provided earlier captures the entire recipe in a single glance, while the FAQ and the “what‑if” toolbox equip you to handle the myriad small variations you will meet in textbooks, exams, or research.
In short, the apparent complexity of (\displaystyle\int\frac{\sqrt{1-\sin^{2}\theta}}{1+\sin\theta},d\theta) is a mirage; behind it lies a tidy combination of algebraic simplification and a well‑chosen substitution. Master these tools, and you will find that most trigonometric integrals become not a chore but a pleasant exercise in pattern recognition Practical, not theoretical..
Happy integrating, and may your future calculations be ever‑clean and ever‑compact.
