CUET UG Mathematics Booster Test 1 - Methods of Integration -

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CUET UG Mathematics Booster Test 1 - Methods of Integration

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Mathematics Booster Test 1 - Methods of Integration

1 / 20

QUESTION 1 OF 20

The area under the graph of
\(y=e^{x^{2}}\)
from 0 to 1 cannot be calculated using the Fundamental Theorem of Calculus with simple inspection because:

Answer Review

1 The area under the graph of
\(y=e^{x^{2}}\)
from 0 to 1 cannot be calculated using the Fundamental Theorem of Calculus with simple inspection because:

A The area is strictly infinite.

B No elementary anti-derivative exists for \(e^{x^{2}}\).

C The function is discontinuous on [0,1].

D The function is always negative.

💡 Short Explanation

FTC requires an antiderivative. \(e^{x^{2}}\)has no elementary antiderivative. Simple inspection therefore fails.

✅ Correct Answer Explanation

The Fundamental Theorem of Calculus evaluates definite integrals using antiderivatives. However, \(e^{x^{2}}\)does not possess an elementary antiderivative expressible through standard algebraic, trigonometric, logarithmic, or exponential functions. Thus simple inspection cannot be used. Numerical methods or special functions are required, making Option B correct.

❓ Why Other Options Are Incorrect

Option A → The area on [0,1] is finite.
Option C → \(e^{x^{2}}\)is continuous everywhere.
Option D → \(e^{x^{2}}\)is always positive.

🎯 Strategy

Used: Elimination

Application: Remove statements contradicting known properties of \(e^{x^{2}}\).

Final Logic: Lack of an elementary antiderivative limits inspection.

🧠 Memory Trick

"No antiderivative → No inspection."

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

2 To find the continuous moving average of
\(f(x)=xe^{x}\)
on an interval [a,b], one must evaluate
\(\int xe^{x} dx.\)
Why does substitution fail here?

A The derivative of \(e^{x}\)or \(x\) does not fully match the remaining part of the integrand, requiring integration by parts.

B Exponential functions do not have moving averages.

C The interval [a,b] contains a singularity.

D The integral evaluates to zero natively.

💡 Short Explanation

Product of two functions appears. No suitable substitution exists. Integration by parts is required.

✅ Correct Answer Explanation

For substitution, one factor should closely resemble the derivative of another. In \(xe^{x}\), neither \(x\) nor \(e^{x}\)produces the remaining factor through differentiation. Therefore substitution is ineffective. Integration by parts is the standard NCERT technique, making Option A correct.

❓ Why Other Options Are Incorrect

Option B → Moving averages can involve exponential functions.
Option C → \(xe^{x}\)has no singularity.
Option D → The integral is non-zero.

🎯 Strategy

Used: Elimination

Application: Identify why substitution requirements are not satisfied.

Final Logic: Product form suggests integration by parts.

🧠 Memory Trick

"Product? Think Parts."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Need for Integration by Parts

3 Evaluate
\(\int x\sqrt{x+2} dx\)
using the change of variable \(t=x+2\).

A \(\frac{2}{5}(x+2)^{5/2}+C\)

B \(\frac{2}{3}(x+2)^{3/2}+C\)

C \(\frac{2}{5}(x+2)^{5/2}-\frac{4}{3}(x+2)^{3/2}+C\)

D \(\frac{1}{2}(x+2)^{2}+C\)

💡 Short Explanation

Let \(t=x+2\). Replace \(x=t-2\). Integrate power terms separately.

✅ Correct Answer Explanation

Using \(t=x+2\), we get \(x=t-2\) and \(dx=dt\). \(\int (t-2)t^{1/2}dt=\int (t^{3/2}-2t^{1/2})dt=\frac{2}{5}t^{5/2}-\frac{4}{3}t^{3/2}+C.\) Substituting back gives Option C.

❓ Why Other Options Are Incorrect

Option A → Omits the \(-2t^{1/2}\)contribution.
Option B → Integrates only one term.
Option D → Does not arise from power-rule integration.

🎯 Strategy

Used: Substitution

Application: Convert the radical expression into polynomial powers.

Final Logic: Expand after substitution and integrate term-wise.

🧠 Memory Trick

"Shift variable, then expand."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Change of Variable

4 A cumulative distribution function involves finding
\(F(x)=\int \frac{2x}{1+x^{2}} dx.\)
Using \(t=1+x^{2}\)and properly transforming \(dx\), what is \(F(x)\)?

A \({tan⁡}^{-1}(x)+C\)

B \(ln⁡(1+x^{2})+C\)

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

D \(ln⁡(2x)+C\)

💡 Short Explanation

Let \(t=1+x^{2}\). Then \(dt=2x dx\). Integral becomes logarithmic.

✅ Correct Answer Explanation

Using \(t=1+x^{2},dt=2x dx,\) the integral becomes \(\int \frac{dt}{t}=ln⁡∣t∣+C.\) Since \(1+x^{2}>0\), \(F(x)=ln⁡(1+x^{2})+C.\) Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Integral of \(1/(1+x^{2})\).
Option C → Differentiates incorrectly.
Option D → Uses an invalid logarithmic argument.

🎯 Strategy

Used: Substitution

Application: Match denominator with its derivative.

Final Logic: Integral reduces to \(\int dt/t\).

🧠 Memory Trick

"Derivative over function → Log."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Logarithmic Integrals

5 An object's acceleration is given by
\(\vec{a}(t)=(sin⁡tcos⁡t)\hat{j}.\)
Finding the velocity
\(\vec{v}(t)=\int \vec{a}(t) dt\)
requires a substitution. Let \(u=sin⁡t\). The velocity vector is:

A \(\frac{{cos⁡}^{2}t}{2}\hat{j}+\vec{C}\)

B \(-\frac{{sin⁡}^{2}t}{2}\hat{j}+\vec{C}\)

C \(sin⁡t\hat{j}+\vec{C}\)

D \(\frac{{sin⁡}^{2}t}{2}\hat{j}+\vec{C}\)

💡 Short Explanation

Let \(u=sin⁡t\). Then \(du=cos⁡t dt\). Integrate \(u du\).

✅ Correct Answer Explanation

Using \(u=sin⁡t,du=cos⁡t dt,\) we obtain \(\int sin⁡tcos⁡t dt=\int u du=\frac{u^{2}}{2}+C.\) Hence \(\vec{v}(t)=\frac{{sin⁡}^{2}t}{2}\hat{j}+\vec{C}.\) Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Wrong trigonometric function.
Option B → Incorrect sign.
Option C → Derivative gives \(\cos\,t\), not \(sin⁡tcos⁡t\).

🎯 Strategy

Used: Substitution

Application: Replace one trigonometric factor using a new variable.

Final Logic: Integral becomes \(\int u du\).

🧠 Memory Trick

"Sin–Cos pair → Substitute Sin."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Trigonometric Substitution

6 Put the steps in order to integrate
\(\int {sin⁡}^{3}x dx:\)
1. Put \(t=cos⁡x\), so \(dt=-sin⁡x dx\).
2. Rewrite as \(\int {sin⁡}^{2}xsin⁡x dx\).
3. Substitute \({sin⁡}^{2}x=1-{cos⁡}^{2}x\).
4. Integrate \(-\int (1-t^{2}) dt\) and replace \(t\) with \(\cos\,x\).

A 1, 2, 3, 4

B 2, 3, 1, 4

C 3, 2, 1, 4

D 1, 3, 2, 4

💡 Short Explanation

Separate one sine factor. Use identity. Apply substitution.

✅ Correct Answer Explanation

The standard method is: 2 → Write \({sin⁡}^{3}x={sin⁡}^{2}xsin⁡x\) 3 → Replace \({sin⁡}^{2}x\) 1 → Substitute \(t=cos⁡x\) 4 → Integrate and back-substitute. Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Substitution occurs too early.
Option C → Identity requires rewriting first.
Option D → Wrong sequence of simplification.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Follow the standard odd-power sine procedure.

Final Logic: Rewrite → Identity → Substitute → Integrate.

🧠 Memory Trick

"Odd sine: save one sine."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Trigonometric Integrals

7 To evaluate
\(\int \frac{\sin\,x}{1+{cos⁡}^{2}x} dx,\)
which substitution is appropriate and what is the final functional form?
I. Substitute \(t=cos⁡x\)
II. The result involves an inverse tangent function
III. The result involves a natural logarithm

A I only

B II only

C I and III

D I and II

💡 Short Explanation

Let \(t=cos⁡x\). Integral becomes \(1/(1+t^{2})\). Result is inverse tangent.

✅ Correct Answer Explanation

Using \(t=cos⁡x,dt=-sin⁡x dx,\) the integral becomes \(-\int \frac{dt}{1+t^{2}}=-{tan⁡}^{-1}(t)+C.\) Substituting back gives \(-{tan⁡}^{-1}(cos⁡x)+C.\) Therefore statements I and II are correct.

❓ Why Other Options Are Incorrect

Option A → Omits the inverse tangent result.
Option B → Ignores the required substitution.
Option C → Natural logarithm does not arise.

🎯 Strategy

Used: Substitution

Application: Convert denominator into the standard \(1+t^{2}\)form.

Final Logic: Integral matches inverse tangent.

🧠 Memory Trick

"1+t² → Tan⁻¹."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Trigonometric Substitution

8 Calculate the exact numerical value of the definite integral
\(\int_{0}^{\pi /4}\,tan⁡x {sec⁡}^{2}x dx\)
by using the substitution \(t=tan⁡x\).

A 1

B 2

C 1/2

D π/4

💡 Short Explanation

Let \(t=tan⁡x\). Then \(dt={sec⁡}^{2}x dx\). Evaluate definite limits.

✅ Correct Answer Explanation

The limits become \(x=0\Rightarrow t=0,x=\pi /4\Rightarrow t=1.\) Hence \(\int_{0}^{1}\,t dt={\left[\frac{t^{2}}{2}\right]}_{0}^{1}=\frac{1}{2}.\) Thus Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Twice the actual value.
Option B → Far larger than the integral.
Option D → Confuses area with interval length.

🎯 Strategy

Used: Substitution

Application: Transform the integral into a simple polynomial.

Final Logic: Integral becomes \(\int_{0}^{1}\,t dt\).

🧠 Memory Trick

"Tan + Sec² → t dt."

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

9 Assertion (A):
\(\int tan⁡x dx={sec⁡}^{2}x+C.\)
Reason (R): The derivative of tan x is \(ln⁡∣sec⁡x∣\).

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Integral formula is incorrect. Derivative statement is incorrect. Both assertions fail.

✅ Correct Answer Explanation

The correct integral is \(\int tan⁡x dx=ln⁡∣sec⁡x∣+C.\) Also, \(\frac{d}{dx}(tan⁡x)={sec⁡}^{2}x.\) Thus the assertion and reason are both false. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Assertion is false.
Option C → Both statements are not true.
Option D → Reason is also false.

🎯 Strategy

Used: Elimination

Application: Compare statements with standard derivative and integral formulas.

Final Logic: Both calculus identities are incorrect.

🧠 Memory Trick

"Tan differentiates to Sec², integrates to Log Sec."

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

10 Which of the following statements is INCORRECT regarding the integral of cot x?

A \(\int cot⁡x dx=ln⁡∣sin⁡x∣+C\)

B \(\int cot⁡x dx=-{csc⁡}^{2}x+C\)

C The substitution \(t=sin⁡x\) is used to derive the standard result

D The domain must not include multiples of \(\pi\) where \(sin⁡x=0\)

💡 Short Explanation

Standard result is logarithmic. Option B is a derivative expression. It is not an antiderivative.

✅ Correct Answer Explanation

The standard formula is \(\int cot⁡x dx=ln⁡∣sin⁡x∣+C.\) Using \(t=sin⁡x\) yields this result directly. Since \(sin⁡x=0\) causes discontinuities, such points are excluded. Option B is incorrect because \(-{csc⁡}^{2}x\) is related to differentiation, not integration.

❓ Why Other Options Are Incorrect

Option A → Correct standard formula.
Option C → Correct derivation method.
Option D → Correct domain restriction.

🎯 Strategy

Used: Option Grouping

Application: Compare each statement with the standard cotangent integral.

Final Logic: Only Option B contradicts the known result.

🧠 Memory Trick

"Cot gives Log Sin."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Integral of cot x

11 The integral
is derived by multiplying the numerator and denominator by (sec x+tan x). The standard result is:

A ln|sec x−tan x|+C

B sec x tan x+C

C ln|sec x+tan x|+C

D (sec²x)/2+C

💡 Short Explanation

Use the standard sec x integration trick. Multiply by (sec x+tan x). Convert to logarithmic form.

✅ Correct Answer Explanation

The classical derivation gives \(\int sec⁡x dx=ln⁡∣sec⁡x+tan⁡x∣+C.\) The numerator becomes the derivative of \(\left(sec⁡x+tan⁡x\right)\), leading to a logarithmic integral. Therefore Option C is correct. Options A, B, and D do not represent the standard antiderivative of sec x.

❓ Why Other Options Are Incorrect

Option A → Related by logarithmic identities but not the standard NCERT result.
Option B → Differentiation gives extra terms.
Option D → Derivative does not equal sec x.

🎯 Strategy

Used: Inspection

Application: Recall the standard trigonometric integral formula.

Final Logic: ∫sec x dx = ln|sec x+tan x|+C.

🧠 Memory Trick

"Sec joins Tan inside Log."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Integral of sec x

12 The integral of csc x is given by
∫csc x dx.
The correct standard mathematical result is:

A -ln|csc x+cot x|+C

B ln|sin x|+C

C csc x cot x+C

D ln|csc x-cot x|+C

💡 Short Explanation

Use the standard csc x integration formula. Convert into logarithmic form. Differentiate to verify.

✅ Correct Answer Explanation

The standard result is \(\int csc⁡x dx=ln⁡∣csc⁡x-cot⁡x∣+C.\) Differentiating the expression reproduces csc x. Therefore Option D is correct. This is the standard NCERT form for the integral of cosecant.

❓ Why Other Options Are Incorrect

Option A → Equivalent up to a constant/sign identity but not the standard listed result.
Option B → Integral of cot x.
Option C → Not an antiderivative of csc x.

🎯 Strategy

Used: Option Grouping

Application: Compare with standard trigonometric integral forms.

Final Logic: Cosecant integrates to a logarithmic expression.

🧠 Memory Trick

"Cosec gives Log(Cosec−Cot)."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Integral of csc x

13 The area of the region modeled by the integral
∫x³/(x⁴+1) dx
evaluates to:

A ln(x⁴+1)+C

B 1/3 ln(x⁴+1)+C

C 1/4 ln(x⁴+1)+C

D 4ln(x⁴+1)+C

💡 Short Explanation

Let t=x⁴+1. Then dt=4x³dx. Integral becomes logarithmic.

✅ Correct Answer Explanation

Using \(t=x^{4}+1,dt=4x^{3}dx,\) the integral becomes \(\frac{1}{4}\int \frac{dt}{t}=\frac{1}{4}ln⁡∣t∣+C.\) Substituting back gives \(\frac{1}{4}ln⁡(x^{4}+1)+C.\) Thus Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Missing factor 1/4.
Option B → Incorrect coefficient.
Option D → Coefficient is four times too large.

🎯 Strategy

Used: Substitution

Application: Recognize denominator and derivative pattern.

Final Logic: x³dx matches one-fourth of dt.

🧠 Memory Trick

"Derivative of x⁴ gives the quarter."

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

14 Match the radical expression to the correct trigonometric substitution used to simplify it:

List I	List II
1. √(a²−x²)	a. x=a sin θ
2. √(a²+x²)	b. x=a sec θ
3. √(x²−a²)	c. x=a tan θ
4. 1/√x (Algebraic)	d. t=√x

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

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

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

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

💡 Short Explanation

Use standard trigonometric substitutions. Each radical has a preferred form. Match according to identities.

✅ Correct Answer Explanation

For \(\sqrt{a^{2}-x^{2}},\) use \(x=asin⁡\theta\). For \(\sqrt{a^{2}+x^{2}},\) use \(x=atan⁡\theta\). For \(\sqrt{x^{2}-a^{2}},\) use \(x=asec⁡\theta\). For \(1/\sqrt{x}\), use \(t=\sqrt{x}\). Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → First three substitutions are mismatched.
Option C → Does not follow standard radical forms.
Option D → Incorrectly matches algebraic and trigonometric substitutions.

🎯 Strategy

Used: Option Grouping

Application: Recall standard trigonometric substitution identities.

Final Logic: Each radical corresponds to a unique substitution pattern.

🧠 Memory Trick

"Minus–Sin, Plus–Tan, Difference–Sec."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Trigonometric Substitution

15 Assertion (A):
∫2x sin(x²) dx = sin(x²)+C.
Reason (R):
d/dx(-cos(x²)) = 2x sin(x²).

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Differentiate −cos(x²). Obtain the given integrand. Compare with the assertion.

✅ Correct Answer Explanation

Since \(\frac{d}{dx}[-cos⁡(x^{2})]=2xsin⁡(x^{2}),\) the correct integral is \(-cos⁡(x^{2})+C.\) Therefore the assertion is false, while the reason is true. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Reason is true.
Option B → Assertion is false.
Option C → Assertion itself is incorrect.

🎯 Strategy

Used: Elimination

Application: Verify the derivative and compare with the assertion.

Final Logic: Integrand corresponds to −cos(x²), not sin(x²).

🧠 Memory Trick

"Sin integrates to −Cos."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Reverse Chain Rule

16 By recognizing the inner function, evaluate
∫1/(x log x) dx.

A ln|ln x|+C

B (ln x)²/2+C

C xln x+C

D ln|x|+C

💡 Short Explanation

Let t=ln x. Then dt=dx/x. Integral becomes ∫dt/t.

✅ Correct Answer Explanation

Using \(t=ln⁡x,dt=\frac{dx}{x},\) the integral becomes \(\int \frac{dt}{t}=ln⁡∣t∣+C.\) Replacing t by ln x gives \(ln⁡∣ln⁡x∣+C.\) Thus Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Integral of ln x/x.
Option C → Derivative contains additional terms.
Option D → Integral of 1/x.

🎯 Strategy

Used: Substitution

Application: Identify the inner logarithmic function.

Final Logic: Integral reduces to ∫1/t dt.

🧠 Memory Trick

"Log inside Log."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Logarithmic Substitution

17 Evaluate
∫sin3x cos4x dx
using the identity 2sinA cosB = sin(A+B)+sin(A−B).

A (cos7x)/7−cosx+C

B −(cos7x)/7−cosx+C

C −(cos7x)/14+cosx/2+C

D (sin7x)/14+sinx/2+C

💡 Short Explanation

Apply product-to-sum identity. Integrate resulting sine terms. Simplify constants.

✅ Correct Answer Explanation

Using \(2sin⁡3xcos⁡4x=sin⁡7x-sin⁡x,\) we get \(sin⁡3xcos⁡4x=\frac{1}{2}(sin⁡7x-sin⁡x).\) Integrating: \(-\frac{cos⁡7x}{14}+\frac{\cos\,x}{2}+C.\) Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Missing factor 1/2 adjustment.
Option B → Incorrect coefficient of cos x.
Option D → Integrates sine incorrectly.

🎯 Strategy

Used: Substitution

Application: Convert product into a sum before integrating.

Final Logic: Product-to-sum identity simplifies the integral.

🧠 Memory Trick

"Product → Sum → Integrate."

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

18 Sequence the steps to integrate
∫cos x/(cos(x−a)) dx:
1. Substitute to get ∫(cos(t+a))/cos t dt
2. Let x−a=t⇒dx=dt, x=t+a
3. Expand cos(t+a)=cos t cos a−sin t sin a
4. Divide by cos t and integrate to obtain x cos a−sin a ln|sec(x−a)|+C

A 1, 2, 3, 4

B 2, 1, 3, 4

C 2, 3, 1, 4

D 1, 3, 2, 4

💡 Short Explanation

Substitute first. Rewrite in the new variable. Expand and integrate.

✅ Correct Answer Explanation

The correct process is: 2 → Define t=x−a. 1 → Rewrite the integral in t. 3 → Expand cos(t+a). 4 → Simplify and integrate. Therefore Option B gives the correct sequence.

❓ Why Other Options Are Incorrect

Option A → Rewriting occurs before substitution.
Option C → Expansion requires rewriting first.
Option D → Substitution must occur before expansion.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Follow the natural order of substitution and simplification.

Final Logic: Substitute → Rewrite → Expand → Integrate.

🧠 Memory Trick

"Substitute before Expand."

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

📄 Passage

Integration by substitution is a powerful method used to reduce complex integrals into standard forms. According to standard calculus principles, if we have an integral of the form
∫f(g(x))g'(x) dx,
we can substitute g(x)=t, which gives g'(x)dx=dt. This transforms the integral into ∫f(t) dt, which is often much easier to evaluate.
Moreover, this method extends to trigonometric identities, where expressions like sin²x or cos²x are replaced using double-angle formulas before applying substitution or direct integration.

Based on the passage, if an identity is required first, what is the value of
∫cos²x dx?

A x/2+(sin2x)/4+C

B x/2−(sin2x)/4+C

C (cos³x)/3+C

D sin²x+C

💡 Short Explanation

Use the power-reduction identity. Integrate term-wise. Simplify the result.

✅ Correct Answer Explanation

Using \({cos⁡}^{2}x=\frac{1+cos⁡2x}{2},\) we get \(\int {cos⁡}^{2}x dx=\frac{x}{2}+\frac{sin⁡2x}{4}+C.\) Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Sign corresponds to the sin²x result.
Option C → Not the antiderivative of cos²x.
Option D → Differentiation does not return cos²x.

🎯 Strategy

Used: Substitution

Application: Apply the double-angle identity before integrating.

Final Logic: cos²x=(1+cos2x)/2.

🧠 Memory Trick

"Cos-square gives Plus sign."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Power Reduction Formula

📄 Passage

Following the passage's strategy for powers of trigonometric functions, evaluate
∫sin³x dx
at x=π, given that the constant of integration C=0.

A -1/3

B 4/3

C 0

D 2/3

💡 Short Explanation

Rewrite sin³x as sinx(1−cos²x). Integrate using substitution. Evaluate at x=π.

✅ Correct Answer Explanation

\(\int {sin⁡}^{3}x dx=\int sin⁡x(1-{cos⁡}^{2}x) dx=\frac{{cos⁡}^{3}x}{3}-cos⁡x+C.\) With C=0 and x=\pi, \(\frac{{\left(-1\right)}^{3}}{3}-(-1)=-\frac{1}{3}+1=\frac{2}{3}.\) Therefore the provided answer is incorrect.

❓ Why Other Options Are Incorrect

Option A → Omits the +1 contribution.
Option B → Arithmetic and evaluation error.
Option C → Integral value at π is not zero.

🎯 Strategy

Used: Substitution

Application: Convert odd sine power into a cosine polynomial.

Final Logic: F(π)=2/3 when C=0.

🧠 Memory Trick

"Odd sine → Save one sine."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Trigonometric Integrals

CUET UG Mathematics Booster Test 1 - Methods of Integration

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