CUET UG Mathematics Booster Test 2 - Methods of Integration -

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CUET UG Mathematics Booster Test 2 - Methods of Integration

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Mathematics Booster Test 2 - Methods of Integration

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QUESTION 1 OF 20

Assertion (A):
\(\int e^{-x^{2}}dx\)
can be easily solved to an exact elementary formula by the method of inspection.
Reason (R): Standard elementary functions do not always have an anti-derivative expressible in finite elementary terms.

Answer Review

1 Assertion (A):
\(\int e^{-x^{2}}dx\)
can be easily solved to an exact elementary formula by the method of inspection.
Reason (R): Standard elementary functions do not always have an anti-derivative expressible in finite elementary terms.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

\(e^{-x^{2}}\)has no elementary anti-derivative. Inspection cannot identify a standard primitive function. Reason correctly explains the limitation.

✅ Correct Answer Explanation

The assertion is false because \(\int e^{-x^{2}}dx\) cannot be expressed using standard elementary functions and is represented through the Error Function. The reason is true since many continuous functions possess anti-derivatives that are not expressible in finite elementary form. Therefore, inspection fails here, making Option D correct.

❓ Why Other Options Are Incorrect

Option A → Reason is true because non-elementary anti-derivatives exist.
Option B → Assertion is false since \(e^{-x^{2}}\)cannot be integrated by simple inspection.
Option C → Assertion is false, so both statements cannot be true together.

🎯 Strategy

Used: Elimination

Application:

Check the truth value of Assertion and Reason separately using standard integration theory.

Final Logic:

Assertion is false while Reason is true.

🧠 Memory Trick

"Gaussian → Error Function."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Limitation of Inspection Method

2 Identify the INCORRECT statement concerning integration methods:

A Some integrals require splitting the function into partial fractions.

B Integration by substitution is the reverse of the chain rule.

C All continuous mathematical functions have anti-derivatives expressible as standard finite elementary functions.

D Integration by parts is derived from the product rule of differentiation.

💡 Short Explanation

Continuous functions have anti-derivatives. Elementary expressions are not always possible. Option C overstates the result.

✅ Correct Answer Explanation

Although every continuous function has an anti-derivative, it need not be expressible through elementary functions. Functions like \(e^{-x^{2}}\)provide counterexamples. Options A, B and D describe standard integration techniques correctly. Hence Option C is the incorrect statement.

❓ Why Other Options Are Incorrect

Option A → Partial fractions are a standard integration technique for rational functions.
Option B → Substitution directly reverses the chain rule.
Option D → Integration by parts follows from the product rule.

🎯 Strategy

Used: Extreme Word Filter

Application:

The word "All" often signals an incorrect universal statement.

Final Logic:

Not every anti-derivative is elementary.

🧠 Memory Trick

"Continuous ≠ Elementary."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Methods of Integration

3 Find the mean value of the continuous function
\(f(x)=\frac{{tan⁡}^{-1}x}{1+x^{2}}\)
over the interval \(\left[0\ ,\ 1\right]\).

A \(\frac{{\pi}^{2}}{16}\)

B \(\frac{{\pi}^{2}}{32}\)

C \(\frac{\pi }{8}\)

D \(\frac{{\pi}^{2}}{8}\)

💡 Short Explanation

Mean value equals definite integral divided by interval length. Use \(t={tan⁡}^{-1}x\). Integrate \(t\) directly.

✅ Correct Answer Explanation

Mean value: \(\frac{1}{1-0}\int_{0}^{1}\,\frac{{tan⁡}^{-1}x}{1+x^{2}}dx\) Let \(t={tan⁡}^{-1}x,dt=\frac{dx}{1+x^{2}}\) Then \(\int_{0}^{\pi /4}\,t dt=\frac{t^{2}}{2}∣_{0}^{\pi /4}=\frac{{\pi}^{2}}{32}\) Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Twice the correct value.
Option C → Missing quadratic dependence on \(\pi\).
Option D → Four times the correct value.

🎯 Strategy

Used: Substitution

Application:

Convert inverse trigonometric expression into a simple variable.

Final Logic:

Integral becomes \(\int t dt\).

🧠 Memory Trick

"tan⁻¹x with 1+x² → Substitute instantly."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Integration by Substitution

4 Determine the area of the region bounded by
\(y=\frac{e^{x}}{e^{2x}+1},x=0 to \infty\)
and the x-axis.

A \(\frac{\pi }{4}\)

B \(\frac{\pi }{2}\)

C 1

D \(\infty\)

💡 Short Explanation

Use \(t=e^{x}\). Integral transforms into inverse tangent form. Evaluate improper limits carefully.

✅ Correct Answer Explanation

\(I=\int_{0}^{\infty }\,\frac{e^{x}}{e^{2x}+1}dx\) Let \(t=e^{x},dt=e^{x}dx\) Then \(I=\int_{1}^{\infty }\,\frac{dt}{1+t^{2}}={tan⁡}^{-1}t∣_{1}^{\infty }=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}\) Thus Option A is correct. The provided answer is incorrect.

❓ Why Other Options Are Incorrect

Option B → Lower limit contribution is ignored.
Option C → Integral evaluates to \(\pi /4\), not 1.
Option D → Integral converges.

🎯 Strategy

Used: Substitution

Application:

Convert exponential expression into a standard rational form.

Final Logic:

Result reduces to inverse tangent evaluation.

🧠 Memory Trick

"eˣ substitution → arctan."

📚 Class XII – NCERT Mathematics Part II – Chapter: Definite Integrals – Topic: Definite Integrals by Substitution

5 Let
\(f(x)=kxe^{-x^{2}},x\in [0,\infty )\)
be a probability density function.
To ensure total probability is 1, find \(k\).

A \(\frac{1}{2}\)

B 1

C 4

D 2

💡 Short Explanation

Total probability equals 1. Integrate density over its domain. Solve for \(k\).

✅ Correct Answer Explanation

\(1=\int_{0}^{\infty }\,kxe^{-x^{2}}dx\) Using \(t=x^{2},dt=2x dx\) gives \(1=\frac{k}{2}\int_{0}^{\infty }\,e^{-t}dt=\frac{k}{2}\) Hence \(k=2\) Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Gives probability \(1/4\).
Option B → Gives probability \(1/2\).
Option C → Gives probability 2.

🎯 Strategy

Used: Substitution

Application:

Transform Gaussian-type density into a simple exponential integral.

Final Logic:

\(k/2=1\Rightarrow k=2\).

🧠 Memory Trick

"PDF area = 1."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Integration by Substitution

6 The force vector field is
\(\vec{F}(x)=\frac{x}{\sqrt{1-x^{2}}}\hat{i}.\)
The work done in moving a particle from \(x=0\) to \(x=\frac{1}{2}\)is:

A \(\frac{\sqrt{3}}{2}\)

B \(1-\frac{\sqrt{3}}{2}\)

C \(\frac{1}{2}\)

D \(1-\frac{1}{\sqrt{2}}\)

💡 Short Explanation

Work equals force integral. Use \(t=1-x^{2}\). Evaluate definite limits.

✅ Correct Answer Explanation

\(W=\int_{0}^{1/2}\,\frac{x}{\sqrt{1-x^{2}}}dx\) Let \(t=1-x^{2},dt=-2x dx\) Then \(W=-\sqrt{t}∣_{0}^{1/2}=1-\frac{\sqrt{3}}{2}\) Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Only the square-root term appears.
Option C → Incorrect evaluation of limits.
Option D → Corresponds to a different upper limit.

🎯 Strategy

Used: Substitution

Application:

Match numerator with derivative of radical expression.

Final Logic:

Work equals \(1-\sqrt{3}/2\).

🧠 Memory Trick

"Inside radical derivative present → Substitute."

📚 Class XII – NCERT Mathematics Part II – Chapter: Definite Integrals – Topic: Evaluation by Substitution

7 Consider the integral
\(\int \frac{\sin\,x}{sin⁡(x-a)}dx.\)
Which methods are effective?
I. Expand \(sin⁡x=sin⁡((x-a)+a)\)
II. Substitute \(t=x-a\)
III. Substitute \(t=sin⁡x\)

A I only

B II and III

C I and II

D III only

💡 Short Explanation

Shifted angle suggests substitution. Expansion simplifies numerator. Direct \(t=sin⁡x\) is ineffective.

✅ Correct Answer Explanation

Since the denominator contains \(\left(x\ −\ a\right)\), substitution \(t=x-a\) is natural. Also, \(sin⁡x=sin⁡((x-a)+a)\) allows angle expansion and simplification. Substituting \(t=sin⁡x\) does not simplify the denominator. Hence statements I and II are correct.

❓ Why Other Options Are Incorrect

Option A → Expansion alone is insufficient.
Option B → Statement III is not useful.
Option D → Ignores denominator structure.

🎯 Strategy

Used: Elimination

Application:

Choose techniques that simplify both numerator and denominator.

Final Logic:

Shifted-angle integrals require expansion and variable shift.

🧠 Memory Trick

"Shifted angle → Shift variable."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Trigonometric Transformations in Integration

8 Given:
\(\int \frac{1}{{sin⁡}^{2}x{cos⁡}^{2}x}dx=tan⁡x-cot⁡x+C,\)
evaluate
\(\int_{\pi /6}^{\pi /3}\,\frac{1}{{sin⁡}^{2}x{cos⁡}^{2}x}dx.\)

A \(\frac{4}{\sqrt{3}}\)

B \(2\sqrt{3}\)

C 0

D \(\frac{1}{\sqrt{3}}\)

💡 Short Explanation

Use the given antiderivative. Apply upper minus lower limit. Simplify exactly.

✅ Correct Answer Explanation

\(F(x)=tan⁡x-cot⁡x\) At \(x=\pi /3\): \(\sqrt{3}-\frac{1}{\sqrt{3}}=\frac{2}{\sqrt{3}}\) At \(x=\pi /6\): \(\frac{1}{\sqrt{3}}-\sqrt{3}=-\frac{2}{\sqrt{3}}\) Thus \(\frac{2}{\sqrt{3}}-\left(\ −\ \frac{2}{\sqrt{3}}\right)=\frac{4}{\sqrt{3}}\) Therefore Option A is correct. The provided answer is incorrect.

❓ Why Other Options Are Incorrect

Option B → Simplification mistake.
Option C → Positive integrand cannot give zero.
Option D → Only one-fourth of the correct value.

🎯 Strategy

Used: Elimination

Application:

Use Fundamental Theorem of Calculus directly.

Final Logic:

Upper value minus lower value gives \(4/\sqrt{3}\).

🧠 Memory Trick

"FTC = Upper − Lower."

📚 Class XII – NCERT Mathematics Part II – Chapter: Definite Integrals – Topic: Evaluation Using Antiderivatives

9 Evaluate:
\(\int tan⁡(2x-3)dx.\)

A \(log⁡∣sec⁡(2x-3)∣+C\)

B \(2log⁡∣sec⁡(2x-3)∣+C\)

C \(\frac{1}{2}log⁡∣sec⁡(2x-3)∣+C\)

D \(-\frac{1}{2}log⁡∣cos⁡(2x-3)∣+C\)

💡 Short Explanation

Use linear substitution. Apply standard \(\int tan⁡x dx\). Divide by derivative of inner function.

✅ Correct Answer Explanation

Let \(t=2x-3,dt=2dx\) Then \(\int tan⁡(2x-3)dx=\frac{1}{2}\int tan⁡t dt=\frac{1}{2}log⁡∣sec⁡t∣+C=\frac{1}{2}log⁡∣sec⁡(2x-3)∣+C\) Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Missing chain-rule factor.
Option B → Incorrect multiplier.
Option D → Equivalent form mathematically, but Option C is the standard answer.

🎯 Strategy

Used: Substitution

Application:

Remove the linear expression first.

Final Logic:

Divide by derivative 2.

🧠 Memory Trick

"Inside derivative ⇒ Divide."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Standard Trigonometric Integrals

10 Match the standard trigonometric integrals with their logarithmic results:

List I	List II
1. \(\int tan⁡x dx\)	a. \(log⁡∣sec⁡x∣+C\)
2. \(\int cot⁡x dx\)	b. \(log⁡∣sin⁡x∣+C\)
3. \(\int sec⁡x dx\)	c. \(log⁡∣sec⁡x+tan⁡x∣+C\)
4. \(\int csc⁡x dx\)	d. \(log⁡∣csc⁡x-cot⁡x∣+C\)

A 1-b, 2-a, 3-d, 4-c

B 1-a, 2-c, 3-b, 4-d

C 1-c, 2-d, 3-a, 4-b

D 1-a, 2-b, 3-c, 4-d

💡 Short Explanation

Recall four standard logarithmic integrals. Match each trigonometric function carefully. Use formula-based elimination.

✅ Correct Answer Explanation

Standard results are: \(\int tan⁡x dx=log⁡∣sec⁡x∣+C\int cot⁡x dx=log⁡∣sin⁡x∣+C\int sec⁡x dx=log⁡∣sec⁡x+tan⁡x∣+C\int csc⁡x dx=log⁡∣csc⁡x-cot⁡x∣+C\) Thus the correct matching is Option D.

❓ Why Other Options Are Incorrect

Option A → Several logarithmic results are interchanged.
Option B → Cotangent and secant formulas are mismatched.
Option C → Nearly all pairings are incorrect.

🎯 Strategy

Used: Option Grouping

Application:

Recall each standard integral independently and match.

Final Logic:

Only Option D matches all four standard formulas.

🧠 Memory Trick

"Tan–Sec, Cot–Sin, Sec–Sec+Tan, Cosec–Cosec−Cot."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Standard Integrals

11 Utilizing advanced techniques (such as integration by parts), evaluate:
\(\int {sec⁡}^{3}x dx\)

A \(\frac{1}{2}(sec⁡xtan⁡x+log⁡∣sec⁡x+tan⁡x∣)+C\)

B \(sec⁡xtan⁡x-log⁡∣sec⁡x+tan⁡x∣+C\)

C \(\frac{{sec⁡}^{4}x}{4}+C\)

D \(\frac{1}{2}{sec⁡}^{2}x+C\)

💡 Short Explanation

Use integration by parts. Reduce \({sec⁡}^{3}x\) to \(\sec\,x\). Apply the standard \(\int sec⁡x dx\) result.

✅ Correct Answer Explanation

The standard reduction formula gives \(\int {sec⁡}^{3}x dx=\frac{1}{2}sec⁡xtan⁡x+\frac{1}{2}\int sec⁡x dx.\) Using \(\int sec⁡x dx=log⁡∣sec⁡x+tan⁡x∣,\) we obtain \(\frac{1}{2}(sec⁡xtan⁡x+log⁡∣sec⁡x+tan⁡x∣)+C.\) Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect coefficient and sign for logarithmic term.
Option C → Differentiation gives \({sec⁡}^{3}xtan⁡x\), not \({sec⁡}^{3}x\).
Option D → Differentiation gives \({sec⁡}^{2}xtan⁡x\).

🎯 Strategy

Used: Elimination

Application:

Recall the standard reduction result for \(\int {sec⁡}^{3}x dx\).

Final Logic:

Only Option A matches the known formula.

🧠 Memory Trick

"Sec³ = Half (SecTan + Log)."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Integration by Parts

12 The standard result
\(\int csc⁡x dx=log⁡∣csc⁡x-cot⁡x∣+C\)
can be expressed using half-angle identities. Which equivalent form is correct?

A \(log⁡∣cos⁡(x/2)∣+C\)

B \(log⁡∣tan⁡(x/2)∣+C\)

C \(log⁡∣cot⁡(x/2)∣+C\)

D \(log⁡∣sin⁡(x/2)∣+C\)

💡 Short Explanation

Use half-angle identities. \(csc⁡x-cot⁡x=tan⁡(x/2)\). Replace inside logarithm.

✅ Correct Answer Explanation

Using the identity \(csc⁡x-cot⁡x=tan⁡\frac{x}{2},\) the standard result becomes \(\int csc⁡x dx=log⁡∣tan⁡\frac{x}{2}∣+C.\) Therefore Option B is the equivalent half-angle form.

❓ Why Other Options Are Incorrect

Option A → Does not satisfy the identity.
Option C → Reciprocal of the correct expression.
Option D → Missing the tangent ratio structure.

🎯 Strategy

Used: Option Grouping

Application:

Compare each option with the half-angle identity.

Final Logic:

\(csc⁡x-cot⁡x=tan⁡(x/2)\).

🧠 Memory Trick

"Cosec−Cot = Tan Half."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Standard Trigonometric Integrals

13 Arrange the correct sequence to evaluate
\(\int x\sqrt{x+a} dx.\)
1. Multiply to obtain \((t-a)\sqrt{t}=t^{3/2}-at^{1/2}\)
2. Differentiate to get \(dx=dt\) and express \(x=t-a\)
3. Put \(x+a=t\)
4. Integrate using the power rule

A 1, 2, 3, 4

B 3, 1, 2, 4

C 2, 4, 1, 3

D 3, 2, 1, 4

💡 Short Explanation

First substitute. Express \(x\) and \(dx\) in \(t\). Expand and integrate.

✅ Correct Answer Explanation

Correct order: 3 → Put \(x+a=t\) 2 → Obtain \(x=t-a, dx=dt\) 1 → Rewrite as \(t^{3/2}-at^{1/2}\) 4 → Apply the power rule. Thus Option D gives the proper substitution procedure.

❓ Why Other Options Are Incorrect

Option A → Expansion occurs before substitution.
Option B → Algebraic expansion precedes variable transformation.
Option C → Integration cannot occur before simplification.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Follow the natural order of substitution and simplification.

Final Logic:

Substitute → Transform → Expand → Integrate.

🧠 Memory Trick

"SETI = Substitute, Express, Transform, Integrate."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Integration by Substitution

14 Evaluate:
\(\int \frac{dx}{x-\sqrt{x}}\)
using the substitution \(t=\sqrt{x}\).

A \(2log⁡∣\sqrt{x}-1∣+C\)

B \(log⁡∣x-\sqrt{x}∣+C\)

C \(2\sqrt{x}+log⁡∣\sqrt{x}∣+C\)

D \(\frac{1}{2}log⁡∣\sqrt{x}-1∣+C\)

💡 Short Explanation

Let \(t=\sqrt{x}\). Convert to rational form. Integrate \(2/(t-1)\).

✅ Correct Answer Explanation

Let \(t=\sqrt{x},x=t^{2},dx=2t dt.\) Then \(\int \frac{dx}{x-\sqrt{x}}=\int \frac{2t dt}{t^{2}-t}=\int \frac{2 dt}{t-1}.\) Hence \(2log⁡∣t-1∣+C=2log⁡∣\sqrt{x}-1∣+C.\) Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Not obtained after substitution.
Option C → Extra algebraic term appears incorrectly.
Option D → Incorrect coefficient.

🎯 Strategy

Used: Substitution

Application:

Transform the radical denominator into a rational expression.

Final Logic:

Integral reduces to \(2\int \frac{dt}{t-1}\).

🧠 Memory Trick

"√x present? Put \(t=\sqrt{x}\)."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Algebraic Substitution

15 Evaluate the integral:
\(\int \frac{2x{sin⁡}^{-1}(x^{2})}{\sqrt{1-x^{4}}} dx.\)

A \({sin⁡}^{-1}(x^{2})+C\)

B \(({sin⁡}^{-1}(x^{2}))^{-1}+C\)

C \(\frac{{\left({sin⁡}^{-1}(x^{2})\right)}^{2}}{2}+C\)

D \(\sqrt{1-x^{4}}+C\)

💡 Short Explanation

Identify inverse trigonometric composite. Use chain-rule reversal. Integrate \(u du\).

✅ Correct Answer Explanation

Let \(u={sin⁡}^{-1}(x^{2}).\) Then \(du=\frac{2x}{\sqrt{1-x^{4}}}dx.\) Thus \(\int u du=\frac{u^{2}}{2}+C=\frac{{\left({sin⁡}^{-1}(x^{2})\right)}^{2}}{2}+C.\) Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Missing squaring effect.
Option B → Inverse reciprocal operation is irrelevant.
Option D → Not related to the substitution result.

🎯 Strategy

Used: Substitution

Application:

Recognize the derivative of \({sin⁡}^{-1}(x^{2})\).

Final Logic:

Integral becomes \(\int u du\).

🧠 Memory Trick

"Function × Derivative → Half Square."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Composite Functions and Substitution

16 Evaluate:
\(\int \frac{sin⁡({tan⁡}^{-1}x)}{1+x^{2}} dx.\)

A \(cos⁡({tan⁡}^{-1}x)+C\)

B \(-cos⁡({tan⁡}^{-1}x)+C\)

C \(sin⁡({tan⁡}^{-1}x)+C\)

D \(log⁡∣1+x^{2}∣+C\)

💡 Short Explanation

Let \(t={tan⁡}^{-1}x\). Then \(dt=\frac{dx}{1+x^{2}}\). Integrate \(\sin\,t\).

✅ Correct Answer Explanation

Using \(t={tan⁡}^{-1}x,\) we obtain \(dt=\frac{dx}{1+x^{2}}.\) Hence \(\int sin⁡t dt=-cos⁡t+C.\) Substituting back, \(-cos⁡({tan⁡}^{-1}x)+C.\) Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Wrong sign.
Option C → Derivative produces cosine.
Option D → Integral is not logarithmic.

🎯 Strategy

Used: Substitution

Application:

Recognize inverse tangent and its derivative.

Final Logic:

Integral reduces to \(\int sin⁡t dt\).

🧠 Memory Trick

"arctan + \(1+x^{2}\)⇒ substitute."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Substitution Method

17 Evaluate:
\(\int {cos⁡}^{4}(2x) dx.\)

A \(\frac{3x}{8}+\frac{sin⁡4x}{4}+\frac{sin⁡8x}{32}+C\)

B \(\frac{x}{4}+\frac{sin⁡4x}{8}+C\)

C \(\frac{{cos⁡}^{5}(2x)}{10}+C\)

D \(\frac{3x}{8}+\frac{sin⁡4x}{8}+\frac{sin⁡8x}{64}+C\)

💡 Short Explanation

Use power-reduction identities twice. Convert powers into multiple angles. Integrate term-wise.

✅ Correct Answer Explanation

Using \({cos⁡}^{4}\theta =\frac{3+4cos⁡2\theta +cos⁡4\theta }{8},\) with \(\theta =2x\), \({cos⁡}^{4}(2x)=\frac{3}{8}+\frac{1}{2}cos⁡4x+\frac{1}{8}cos⁡8x.\) Integrating: \(\frac{3x}{8}+\frac{sin⁡4x}{8}+\frac{sin⁡8x}{64}+C.\) Thus Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Coefficients of sine terms are doubled.
Option B → Ignores higher-angle contribution.
Option C → Differentiation does not produce \({cos⁡}^{4}(2x)\).

🎯 Strategy

Used: Substitution

Application:

Apply power-reduction identities before integrating.

Final Logic:

Reduce powers to linear trigonometric terms.

🧠 Memory Trick

"Cos⁴ → 3,4,1 over 8."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Trigonometric Integrals Using Identities

18 Simplify and evaluate:
\(\int \frac{cos⁡2x-cos⁡2\alpha }{cos⁡x-cos⁡\alpha } dx.\)

A \(2(sin⁡x-xcos⁡\alpha )+C\)

B \(2(sin⁡x+xcos⁡\alpha )+C\)

C \(sin⁡x+cos⁡\alpha +C\)

D \(2xsin⁡\alpha +cos⁡x+C\)

💡 Short Explanation

Use factorization of cosine difference. Simplify the ratio. Integrate the resulting expression.

✅ Correct Answer Explanation

Using \(cos⁡2x-cos⁡2\alpha =2(cos⁡x-cos⁡\alpha )(cos⁡x+cos⁡\alpha ),\) the integrand simplifies to \(2(cos⁡x+cos⁡\alpha ).\) Integrating, \(2sin⁡x+2xcos⁡\alpha +C=2(sin⁡x+xcos⁡\alpha )+C.\) Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Wrong sign before \(xcos⁡\alpha\).
Option C → Missing factor 2 and \(x\)-term.
Option D → Incorrect trigonometric constant.

🎯 Strategy

Used: Elimination

Application:

Factor the numerator before integration.

Final Logic:

Cancellation leads to a simple integrand.

🧠 Memory Trick

"Factor first, integrate later."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Trigonometric Simplification

19 Evaluate:
\(\int \frac{e^{x}(1+x)}{{cos⁡}^{2}(xe^{x})} dx.\)

A \(tan⁡(xe^{x})+C\)

B \(sec⁡(xe^{x})+C\)

C \(cot⁡(xe^{x})+C\)

D \(-tan⁡(xe^{x})+C\)

💡 Short Explanation

Identify inner function \(xe^{x}\). Derivative appears completely. Reverse chain rule applies.

✅ Correct Answer Explanation

Let \(u=xe^{x}.\) Then \(du=e^{x}(1+x)dx.\) Hence \(\int {sec⁡}^{2}u du=tan⁡u+C.\) Substituting back, \(tan⁡(xe^{x})+C.\) Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Derivative gives \(sec⁡utan⁡u\).
Option C → Derivative gives \(-{csc⁡}^{2}u\).
Option D → Wrong sign.

🎯 Strategy

Used: Substitution

Application:

Recognize complete derivative of the inner function.

Final Logic:

Integral becomes \(\int {sec⁡}^{2}u du\).

🧠 Memory Trick

"Derivative present ⇒ Reverse chain rule."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Integration by Substitution

20 For \(0<x<\pi /2\), evaluate:
\(\int \frac{sin⁡x+cos⁡x}{\sqrt{1+sin⁡2x}} dx.\)

A \(sin⁡x-cos⁡x+C\)

B \(log⁡∣sin⁡x+cos⁡x∣+C\)

C \(x+C\)

D \(1+C\)

💡 Short Explanation

Simplify \(1+sin⁡2x\). Use positivity in the given interval. Integrand becomes 1.

✅ Correct Answer Explanation

Since \(1+sin⁡2x=(sin⁡x+cos⁡x)^{2},\) and \(0<x<\pi /2\), \(sin⁡x+cos⁡x>0.\) Therefore \(\sqrt{1+sin⁡2x}=sin⁡x+cos⁡x.\) The integrand becomes \(1.\) Hence \(\int 1 dx=x+C.\) Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Derivative is \(cos⁡x+sin⁡x\).
Option B → Logarithm is unnecessary after simplification.
Option D → Constant cannot be an antiderivative of 1.

🎯 Strategy

Used: Elimination

Application:

Use the given interval to simplify the square root exactly.

Final Logic:

Integrand reduces to 1.

🧠 Memory Trick

"1+sin2x = (sin+cos)²."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Trigonometric Identities in Integration

CUET UG Mathematics Booster Test 2 - Methods of Integration

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