JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Definite Integral of a Rational Trigonometric Function

Q: Why not multiply numerator and denominator of I_1 by sin x instead of half-angle substitution?

Multiplying by sin x gives (2sin x)/(sin^2 x(1+cos x)) = (2sin x)/((1-cos^2 x)(1+cos x)) = (2sin x)/((1-cos x)(1+cos x)^2). Substituting u = cos x leads to partial fractions with a repeated factor (1+u)^2, which takes longer than the direct t = tan(x/2) substitution.

The value of integral from π/3 to π/2 of ((2 + 3 sin x))/(sin x(1 + cos x)) dx is equal to

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A10/3 - √(3) - log_e √(3)B7/2 - √(3) - log_e √(3)C-2 + 3√(3) + log_e √(3)D10/3 - √(3) + log_e √(3)

Step-by-step solutionView

Correct answer

The integral evaluates to $\frac{10}{3} - \sqrt{3} + \log_{e} \sqrt{3}$ by splitting the integrand into two simpler trigonometric integrals.

Option analysis

Why each option works or fails

A · 10/3 - √(3) - log_e √(3)

The student computed the terms correctly but introduced a sign error on the logarithmic term when evaluating the bounds for $\log ∣ \tan (x / 2) ∣$ or $\log ∣ (1 - \cos x) / (1 + \cos x) ∣$ . At the upper limit $x = π / 2$ , $\tan (x / 2) = 1$ , giving $\log (1) = 0$ . Subtracting the lower limit at $x = π / 3$ , where $\tan (x / 2) = 1 / \sqrt{3}$ , yields $- (- \log \sqrt{3}) = + \log \sqrt{3}$ .

B · 7/2 - √(3) - log_e √(3)

The student miscalculated the polynomial antiderivative $\int (1 - t)^{- 2} d t$ or made an arithmetic error when evaluating $3 / (1 + \cos x)$ at the limits. For $\int 3 / (1 + \cos x) d x = 3 \tan (x / 2)$ , evaluating from $π / 3$ to $π / 2$ gives $3 (1 - 1 / \sqrt{3}) = 3 - \sqrt{3}$ , which adds to $1 / 3$ to produce $10 / 3$ , not $7 / 2$ .

C · -2 + 3√(3) + log_e √(3)

The student inverted the signs or bounds when applying the substitution $t = \cos x$ or half-angle formulas. Carefully track upper limit minus lower limit: for $\int_{1 / 2}^{0} \frac{- 2}{(1 - t) (1 + t)^{2}} d t = \int_{0}^{1 / 2} \frac{2}{(1 - t) (1 + t)^{2}} d t$ , ensuring the boundary values are not reversed.

D · 10/3 - √(3) + log_e √(3)

This is the correct evaluation. Splitting the integrand into $2 \int_{π / 3}^{π / 2} \frac{d x}{\sin x (1 + \cos x)} + 3 \int_{π / 3}^{π / 2} \frac{d x}{1 + \cos x}$ leads directly to $\frac{10}{3} - \sqrt{3} + \log_{e} \sqrt{3}$ .

Reviewed route

Solution

StepWorking

01Given

Definite integral $I = \int_{π / 3}^{π / 2} \frac{2 + 3 \sin x}{\sin x (1 + \cos x)} d x$ .

02Goal

Evaluate the exact value of $I$ .

03Approach

Split the integrand into two separate integrals: $I_{1} = \int_{π / 3}^{π / 2} \frac{2}{\sin x (1 + \cos x)} d x$ and $I_{2} = \int_{π / 3}^{π / 2} \frac{3}{1 + \cos x} d x$ . Evaluate $I_{1}$ using the half-angle tangent substitution $t = \tan (x / 2)$ , and $I_{2}$ using the identity $1 + \cos x = 2 \cos^{2} (x / 2)$ or multiplying numerator and denominator by $(1 - \cos x)$ .

04Execute

For $I_{1}$ , substitute $t = \tan (x / 2)$ , so $d x = \frac{2}{1 + t^{2}} d t$ , $\sin x = \frac{2 t}{1 + t^{2}}$ , and $1 + \cos x = \frac{2}{1 + t^{2}}$ . The limits transform from $x = π / 3 ⟹ t = 1 / \sqrt{3}$ to $x = π / 2 ⟹ t = 1$ . The integral becomes: $I_{1} = \int_{1 / \sqrt{3}}^{1} \frac{2}{\frac{2 t}{1 + t^{2}} \cdot \frac{2}{1 + t^{2}}} \cdot \frac{2}{1 + t^{2}} d t = \int_{1 / \sqrt{3}}^{1} \frac{1 + t^{2}}{t} d t = \int_{1 / \sqrt{3}}^{1} (\frac{1}{t} + t) d t$ Evaluating: ${[\ln t + \frac{t^{2}}{2}]}_{1 / \sqrt{3}}^{1} = (0 + \frac{1}{2}) - (\ln (1 / \sqrt{3}) + \frac{1}{6}) = \frac{1}{3} + \ln \sqrt{3}$ .

05Execute

For $I_{2}$ , we have: $I_{2} = 3 \int_{π / 3}^{π / 2} \frac{d x}{2 \cos^{2} (x / 2)} = \frac{3}{2} \int_{π / 3}^{π / 2} \sec^{2} (x / 2) d x = 3 {[\tan (x / 2)]}_{π / 3}^{π / 2} = 3 (1 - \frac{1}{\sqrt{3}}) = 3 - \sqrt{3}$ .

06Execute

Sum $I_{1}$ and $I_{2}$ to find $I$ : $I = I_{1} + I_{2} = (\frac{1}{3} + \ln \sqrt{3}) + (3 - \sqrt{3}) = \frac{10}{3} - \sqrt{3} + \log_{e} \sqrt{3}$ .

✓Verify

Check at endpoints and order of magnitude: on $[π / 3, π / 2]$ , $\sin x \in [\sqrt{3} / 2, 1]$ and $\cos x \in [0, 1 / 2]$ . The integrand varies from $(2 + 3 \sqrt{3} / 2) / (\sqrt{3} / 2 \cdot 3 / 2) \approx 4.598 / 1.299 \approx 3.54$ to $5 / 1 = 5$ . The interval length is $π / 6 \approx 0.5236$ . The integral should be approximately $4.27 \times 0.5236 \approx 2.23$ . Our calculated value $10 / 3 - \sqrt{3} + \ln \sqrt{3} \approx 3.333 - 1.732 + 0.549 = 2.15$ , which matches the mean value estimate perfectly.

Hints that build this answer step by step

How should the integral $I = \int_{π / 3}^{π / 2} \frac{2 + 3 \sin x}{\sin x (1 + \cos x)} d x$ be split to simplify integration?
$I = 2 \int_{π / 3}^{π / 2} \frac{d x}{\sin x (1 + \cos x)} + 3 \int_{π / 3}^{π / 2} \frac{d x}{1 + \cos x}$
What is the value of the second term, $I_{2} = 3 \int_{π / 3}^{π / 2} \frac{d x}{1 + \cos x}$ ?
$3 - \sqrt{3}$
Using the substitution $t = \cos x$ (so $d t = - \sin x d x$ ), what is the value of the first term, $I_{1} = 2 \int_{π / 3}^{π / 2} \frac{\sin x d x}{\sin^{2} x (1 + \cos x)}$ ?
$\frac{1}{3} + \log_{e} \sqrt{3}$
What is the total value $I = I_{1} + I_{2}$ ?
$\frac{10}{3} - \sqrt{3} + \log_{e} \sqrt{3}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

Quick checks

Students also ask

Why not multiply numerator and denominator of I_1 by sin x instead of half-angle substitution?

Multiplying by $\sin x$ gives $\frac{2 \sin x}{\sin^{2} x (1 + \cos x)} = \frac{2 \sin x}{(1 - \cos^{2} x) (1 + \cos x)} = \frac{2 \sin x}{(1 - \cos x) (1 + \cos x)^{2}}$ . Substituting $u = \cos x$ leads to partial fractions with a repeated factor $(1 + u)^{2}$ , which takes longer than the direct $t = \tan (x / 2)$ substitution.

Answer

The integral evaluates to $\frac{10}{3} - \sqrt{3} + \log_{e} \sqrt{3}$ by splitting the integrand into two simpler trigonometric integrals.

Why each option works or fails

A: 10/3 - √(3) - log_e √(3) - The student computed the terms correctly but introduced a sign error on the logarithmic term when evaluating the bounds for $\log ∣ \tan (x / 2) ∣$ or $\log ∣ (1 - \cos x) / (1 + \cos x) ∣$ . At the upper limit $x = π / 2$ , $\tan (x / 2) = 1$ , giving $\log (1) = 0$ . Subtracting the lower limit at $x = π / 3$ , where $\tan (x / 2) = 1 / \sqrt{3}$ , yields $- (- \log \sqrt{3}) = + \log \sqrt{3}$ .
B: 7/2 - √(3) - log_e √(3) - The student miscalculated the polynomial antiderivative $\int (1 - t)^{- 2} d t$ or made an arithmetic error when evaluating $3 / (1 + \cos x)$ at the limits. For $\int 3 / (1 + \cos x) d x = 3 \tan (x / 2)$ , evaluating from $π / 3$ to $π / 2$ gives $3 (1 - 1 / \sqrt{3}) = 3 - \sqrt{3}$ , which adds to $1 / 3$ to produce $10 / 3$ , not $7 / 2$ .
C: -2 + 3√(3) + log_e √(3) - The student inverted the signs or bounds when applying the substitution $t = \cos x$ or half-angle formulas. Carefully track upper limit minus lower limit: for $\int_{1 / 2}^{0} \frac{- 2}{(1 - t) (1 + t)^{2}} d t = \int_{0}^{1 / 2} \frac{2}{(1 - t) (1 + t)^{2}} d t$ , ensuring the boundary values are not reversed.
D · correct: 10/3 - √(3) + log_e √(3) - This is the correct evaluation. Splitting the integrand into $2 \int_{π / 3}^{π / 2} \frac{d x}{\sin x (1 + \cos x)} + 3 \int_{π / 3}^{π / 2} \frac{d x}{1 + \cos x}$ leads directly to $\frac{10}{3} - \sqrt{3} + \log_{e} \sqrt{3}$ .

Step-by-step solution

Given: Definite integral $I = \int_{π / 3}^{π / 2} \frac{2 + 3 \sin x}{\sin x (1 + \cos x)} d x$ .
Goal: Evaluate the exact value of $I$ .
Approach: Split the integrand into two separate integrals: $I_{1} = \int_{π / 3}^{π / 2} \frac{2}{\sin x (1 + \cos x)} d x$ and $I_{2} = \int_{π / 3}^{π / 2} \frac{3}{1 + \cos x} d x$ . Evaluate $I_{1}$ using the half-angle tangent substitution $t = \tan (x / 2)$ , and $I_{2}$ using the identity $1 + \cos x = 2 \cos^{2} (x / 2)$ or multiplying numerator and denominator by $(1 - \cos x)$ .
Execute: For $I_{1}$ , substitute $t = \tan (x / 2)$ , so $d x = \frac{2}{1 + t^{2}} d t$ , $\sin x = \frac{2 t}{1 + t^{2}}$ , and $1 + \cos x = \frac{2}{1 + t^{2}}$ . The limits transform from $x = π / 3 ⟹ t = 1 / \sqrt{3}$ to $x = π / 2 ⟹ t = 1$ . The integral becomes: $I_{1} = \int_{1 / \sqrt{3}}^{1} \frac{2}{\frac{2 t}{1 + t^{2}} \cdot \frac{2}{1 + t^{2}}} \cdot \frac{2}{1 + t^{2}} d t = \int_{1 / \sqrt{3}}^{1} \frac{1 + t^{2}}{t} d t = \int_{1 / \sqrt{3}}^{1} (\frac{1}{t} + t) d t$ Evaluating: ${[\ln t + \frac{t^{2}}{2}]}_{1 / \sqrt{3}}^{1} = (0 + \frac{1}{2}) - (\ln (1 / \sqrt{3}) + \frac{1}{6}) = \frac{1}{3} + \ln \sqrt{3}$ .
Execute: For $I_{2}$ , we have: $I_{2} = 3 \int_{π / 3}^{π / 2} \frac{d x}{2 \cos^{2} (x / 2)} = \frac{3}{2} \int_{π / 3}^{π / 2} \sec^{2} (x / 2) d x = 3 {[\tan (x / 2)]}_{π / 3}^{π / 2} = 3 (1 - \frac{1}{\sqrt{3}}) = 3 - \sqrt{3}$ .
Execute: Sum $I_{1}$ and $I_{2}$ to find $I$ : $I = I_{1} + I_{2} = (\frac{1}{3} + \ln \sqrt{3}) + (3 - \sqrt{3}) = \frac{10}{3} - \sqrt{3} + \log_{e} \sqrt{3}$ .
Verify: Check at endpoints and order of magnitude: on $[π / 3, π / 2]$ , $\sin x \in [\sqrt{3} / 2, 1]$ and $\cos x \in [0, 1 / 2]$ . The integrand varies from $(2 + 3 \sqrt{3} / 2) / (\sqrt{3} / 2 \cdot 3 / 2) \approx 4.598 / 1.299 \approx 3.54$ to $5 / 1 = 5$ . The interval length is $π / 6 \approx 0.5236$ . The integral should be approximately $4.27 \times 0.5236 \approx 2.23$ . Our calculated value $10 / 3 - \sqrt{3} + \ln \sqrt{3} \approx 3.333 - 1.732 + 0.549 = 2.15$ , which matches the mean value estimate perfectly.

Definite Integral of a Rational Trigonometric Function

What feels right?

The integral evaluates to $\frac{10}{3} - \sqrt{3} + \log_{e} \sqrt{3}$ by splitting the integrand into two simpler trigonometric integrals.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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