Integration Techniques JEE 2024: Reflection Errors and Bounds

Q: How do I integrate sin x cos x divided by sin⁴x + cos⁴x?

Use t = tan x to transform the quotient times dx into t/(1 + t⁴) dt. An antiderivative is (1/2) arctan(t²). Between x = 0 and x = π/4, the unweighted integral equals π/8; this does not evaluate the original integral containing the extra factor x.

Q: Is this integral a verified JEE 2024 question?

The supplied metadata does not establish a verified attribution: its year field says 2024, but its date field says 24 January 2023. It also labels the question numerical while giving an option-letter answer without option texts. The original paper and complete answer details are needed before confirming the attribution.

What is wrong with the supplied weighted trigonometric integral solution?

Both claimed answers in this integration techniques jee 2024 entry exceed a provable upper bound. The reflection substitution is valid, but the supplied algebra is not. Conflicting exam metadata also prevents a verified attribution.

The supplied task is to evaluate:

I = \int_{0}^{π / 4} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x .

Two source conflicts need correction:

Year conflict: the year field says 2024, whereas the date field says 2023-01-24, or 24 January 2023. Verify the original paper before using either attribution in a title.
Answer-format conflict: the question is tagged numerical, but its answer is labelled B, and no option texts are supplied.

The steps below verify the reflection identity and reject the claimed answers. They are not a complete evaluation or an endorsed official solution.

How do you apply the reflection substitution correctly?

Reflection changes every occurrence of the integration variable, including those inside the trigonometric functions. The supplied method’s first failure is its claim that the transformed denominator equals the original one. Expanding that denominator exposes the error.

Set:

a = \frac{π}{4}, g (x) = \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x}, I = \int_{0}^{a} x g (x) d x .

Make the substitution and track both limits:

u = a - x, d u = - d x,

x = 0 \Rightarrow u = a, x = a \Rightarrow u = 0 .

Reverse the transformed limits:

I = \int_{a}^{0} (a - u) g (a - u) (- d u) = \int_{0}^{a} (a - u) g (a - u) d u .

Rename the dummy variable:

I = \int_{0}^{a} (a - x) g (a - x) d x .

Transform the numerator using the angle-difference identities:

\sin (a - x) = \frac{\cos x - \sin x}{\sqrt{2}}, \cos (a - x) = \frac{\cos x + \sin x}{\sqrt{2}} .

\sin (a - x) \cos (a - x) = \frac{\cos^{2} x - \sin^{2} x}{2} = \frac{\cos 2 x}{2} .

Expand the denominator:

& \sin^{4} (a - x) + \cos^{4} (a - x) & = \frac{(\cos x - \sin x)^{4} + (\cos x + \sin x)^{4}}{4} & = \frac{2 \cos^{4} x + 12 \sin^{2} x \cos^{2} x + 2 \sin^{4} x}{4} & = \frac{1}{2} + 2 \sin^{2} x \cos^{2} x .

The original denominator is different:

\sin^{4} x + \cos^{4} x = 1 - 2 \sin^{2} x \cos^{2} x .

At the lower endpoint:

x = 0 : D_{reflected} = \frac{1}{2}, D_{original} = 1 .

The denominators are not identical. This endpoint check disproves the supplied invariance claim, so later steps relying on it are invalid.

How can you reject the claimed answer without finishing the integral?

The trigonometric quotient lies between zero and one. Therefore, the weighted integral cannot exceed the integral of the variable alone. This bound rejects both supplied answers without requiring an antiderivative.

First, the denominator satisfies:

\sin^{4} x + \cos^{4} x = 1 - 2 \sin^{2} x \cos^{2} x \geq \frac{1}{2} .

On the stated interval: $0 \leq \sin x \cos x \leq \frac{1}{2} .$

Consequently: $0 \leq g (x) \leq 1 .$

The integrand is positive inside the interval, so:

0 < I \leq \int_{0}^{π / 4} x d x = {[\frac{x^{2}}{2}]}_{0}^{π / 4} = \frac{π^{2}}{32} .

Compare the bound with the supplied answers:

Upper bound:

\frac{π^{2}}{32} \approx 0.30843 .

Supplied intermediate answer:

\frac{π}{8} \approx 0.39270 > 0.30843 .

Supplied final claim:

\frac{46}{3} \approx 15.33333 > 0.30843 .

Publication must pause: a complete worked solution following the supplied derivation and endorsing its final answer cannot be written honestly. This is not a rounding issue.

Request the original question, complete answer choices if applicable, and a corrected official solution. Do not invent different limits or replacement constants to make the claimed answer fit.

When can definite-integral symmetry remove the variable factor?

The reflection shortcut applies when the remaining factor, called the kernel, is unchanged by reflection across the interval’s midpoint. A symmetric-looking trigonometric expression is not enough. Here, the unequal endpoint values immediately disprove the required symmetry.

The shortcut uses this condition:

g (a - x) = g (x) ⟹ \int_{0}^{a} x g (x) d x = \frac{a}{2} \int_{0}^{a} g (x) d x .

But the present kernel gives:

g (0) = 0, g (\frac{π}{4}) = 1 .

Reflection exchanges those endpoints, so the required identity fails. Adding the original and reflected integrals gives:

2 I = \int_{0}^{a} [x g (x) + (a - x) g (a - x)] d x .

The combined integrand can be written as:

x [g (x) - g (a - x)] + a g (a - x) .

The variable-dependent terms do not automatically cancel. The bracketed difference must vanish for that cancellation.

Exchanging sine and cosine is not the same as reflecting across this interval’s midpoint. That exchange preserves the original kernel, but corresponds to:

x \mapsto \frac{π}{2} - x, not x \mapsto \frac{π}{4} - x .

The algebraic error can be identified without guessing a distractor. No wrong-option label can be assigned because the options are missing; option-by-option analysis remains blocked until they are supplied.

Which two practice questions separate substitution from valid symmetry?

The first exercise removes the variable weight, making substitution straightforward. The second uses an interval with genuine kernel symmetry, making reflection valid. Both are original related practice, not verified JEE questions.

How do you integrate the same quotient without the variable weight?

Use the tangent substitution to turn the quotient into a rational expression. Its antiderivative then follows from the inverse-tangent derivative, without any symmetry claim.

Question 1: Evaluate

K = \int_{0}^{π / 4} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x .

Use:

t = \tan x, d x = \frac{d t}{1 + t^{2}},

\sin x \cos x = \frac{t}{1 + t^{2}}, \sin^{4} x + \cos^{4} x = \frac{1 + t^{4}}{(1 + t^{2})^{2}} .

The limits become zero and one:

K = \int_{0}^{1} \frac{t}{1 + t^{4}} d t = {[\frac{1}{2} \arctan (t^{2})]}_{0}^{1} = \frac{π}{8} .

How does reflection work when the kernel really is symmetric?

On the interval below, reflection exchanges sine and cosine, preserving their product. Adding the original and reflected integrals therefore removes the variable weight correctly.

Question 2: Evaluate

J = \int_{0}^{π / 2} x \sin x \cos x d x .

Verify the reflected kernel:

a = \frac{π}{2}, \sin (a - x) \cos (a - x) = \cos x \sin x .

Adding the reflected expression gives:

2 J = \frac{π}{2} \int_{0}^{π / 2} \sin x \cos x d x = \frac{π}{2} {[\frac{\sin^{2} x}{2}]}_{0}^{π / 2} = \frac{π}{4} .

J = \frac{π}{8}

Unlike Question 1, this exercise retains the variable weight but uses an interval with genuine reflection symmetry. Before cancelling a variable factor in your next definite integral, write and verify the reflected kernel explicitly.

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

For a worked example of the same idea, see Redox Reactions and Electrochemistry JEE 2017: Solution.

Frequently asked questions

When can symmetry remove x from a definite integral?

For an integral of xg(x) from 0 to a, the reflection shortcut requires g(a − x) = g(x). Under that condition, the integral equals a/2 times the integral of g(x) over the same interval. For the article's kernel on [0, π/4], the endpoint values are 0 and 1, so the required symmetry fails.

How can I reject an integral answer without evaluating it?

Bound the integrand and integrate the simpler bound. In the supplied weighted integral, the trigonometric quotient lies between 0 and 1, giving 0 < I ≤ π²/32 ≈ 0.30843. Both claimed values, π/8 and 46/3, exceed this upper bound and are therefore incorrect.

How do I integrate sin x cos x divided by sin⁴x + cos⁴x?