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Integral Calculus: JEE Main Mathematics Question with Solution

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

Multiplying by gives . Substituting leads to partial fractions with a repeated factor , which takes longer than the direct substitution.

The value of

\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2 + 3 \sin x)}{\sin x(1 + \cos x)} dx

is equal to

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What feels right?

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\frac{10}{3} - \sqrt{3} - \log_e \sqrt{3}

\frac{7}{2} - \sqrt{3} - \log_e \sqrt{3}

-2 + 3\sqrt{3} + \log_e \sqrt{3}

\frac{10}{3} - \sqrt{3} + \log_e \sqrt{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 ·

\frac{10}{3} - \sqrt{3} - \log_e \sqrt{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 $x = \pi/2$ , $\tan(x/2) = 1$ , giving $\log(1) = 0$ . Subtracting the lower limit at $x = \pi/3$ , where $\tan(x/2) = 1/\sqrt{3}$ , yields $-(- \log\sqrt{3}) = +\log\sqrt{3}$ .

B ·

\frac{7}{2} - \sqrt{3} - \log_e \sqrt{3}

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

C ·

-2 + 3\sqrt{3} + \log_e \sqrt{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 $\int_{1/2}^0 \frac{-2}{(1-t)(1+t)^2} dt = \int_0^{1/2} \frac{2}{(1-t)(1+t)^2} dt$ , ensuring the boundary values are not reversed.

D ·

\frac{10}{3} - \sqrt{3} + \log_e \sqrt{3}

None. This is the correct evaluation. Splitting the integrand into $2\int_{\pi/3}^{\pi/2} \frac{dx}{\sin x(1+\cos x)} + 3\int_{\pi/3}^{\pi/2} \frac{dx}{1+\cos x}$ leads directly to $\frac{10}{3} - \sqrt{3} + \log_e \sqrt{3}$ .

Reviewed route

Solution

StepWorking

01given

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

02goal

Evaluate the exact value of $I$ .

03approach

Split the integrand into two separate integrals: $I_1 = \int_{\pi/3}^{\pi/2} \frac{2}{\sin x(1 + \cos x)} \, dx$ and $I_2 = \int_{\pi/3}^{\pi/2} \frac{3}{1 + \cos x} \, dx$ . 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 $dx = \frac{2}{1+t^2} \, dt$ , $\sin x = \frac{2t}{1+t^2}$ , and $1 + \cos x = \frac{2}{1+t^2}$ . The limits transform from $x = \pi/3 \implies t = 1/\sqrt{3}$ to $x = \pi/2 \implies t = 1$ . The integral becomes: $I_1 = \int_{1/\sqrt{3}}^1 \frac{2}{\frac{2t}{1+t^2} \cdot \frac{2}{1+t^2}} \cdot \frac{2}{1+t^2} \, dt = \int_{1/\sqrt{3}}^1 \frac{1 + t^2}{t} \, dt = \int_{1/\sqrt{3}}^1 \left(\frac{1}{t} + t\right) dt$ Evaluating: $\left[\ln t + \frac{t^2}{2}\right]_{1/\sqrt{3}}^1 = \left(0 + \frac{1}{2}\right) - \left(\ln(1/\sqrt{3}) + \frac{1}{6}\right) = \frac{1}{3} + \ln\sqrt{3}$ .

05execute

For $I_2$ , we have: $I_2 = 3 \int_{\pi/3}^{\pi/2} \frac{dx}{2\cos^2(x/2)} = \frac{3}{2} \int_{\pi/3}^{\pi/2} \sec^2(x/2) \, dx = 3 \left[\tan(x/2)\right]_{\pi/3}^{\pi/2} = 3 \left(1 - \frac{1}{\sqrt{3}}\right) = 3 - \sqrt{3}$ .

06execute

Sum $I_1$ and $I_2$ to find $I$ : $I = I_1 + I_2 = \left(\frac{1}{3} + \ln\sqrt{3}\right) + (3 - \sqrt{3}) = \frac{10}{3} - \sqrt{3} + \log_e \sqrt{3}$ .

✓verify

Check at endpoints and order of magnitude: on $[\pi/3, \pi/2]$ , $\sin x \in [\sqrt{3}/2, 1]$ 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 $\pi/6 \approx 0.5236$ . 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 $I = \int_{\pi/3}^{\pi/2} \frac{2 + 3 \sin x}{\sin x(1 + \cos x)} dx$ be split to simplify integration?
$I = 2 \int_{\pi/3}^{\pi/2} \frac{dx}{\sin x(1 + \cos x)} + 3 \int_{\pi/3}^{\pi/2} \frac{dx}{1 + \cos x}$
What is the value of the second term, $I_2 = 3 \int_{\pi/3}^{\pi/2} \frac{dx}{1 + \cos x}$ ?
$3 - \sqrt{3}$
Using the substitution $t = \cos x$ (so $dt = -\sin x dx$ ), what is the value of the first term, $I_1 = 2 \int_{\pi/3}^{\pi/2} \frac{\sin x dx}{\sin^2 x (1 + \cos x)}$ ?
$\frac{1}{3} + \log_e \sqrt{3}$
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
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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: $\frac{10}{3} - \sqrt{3} - \log_e \sqrt{3}$ - 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 $x = \pi/2$ , $\tan(x/2) = 1$ , giving $\log(1) = 0$ . Subtracting the lower limit at $x = \pi/3$ , where $\tan(x/2) = 1/\sqrt{3}$ , yields $-(- \log\sqrt{3}) = +\log\sqrt{3}$ .
B: $\frac{7}{2} - \sqrt{3} - \log_e \sqrt{3}$ - The student miscalculated the polynomial antiderivative $\int (1-t)^{-2} dt$ or made an arithmetic error when evaluating $3/(1+\cos x)$ at the limits. For $\int 3/(1+\cos x) dx = 3\tan(x/2)$ , evaluating from $\pi/3$ to $\pi/2$ gives $3(1 - 1/\sqrt{3}) = 3 - \sqrt{3}$ , which adds to $1/3$ to produce $10/3$ , not $7/2$ .
C: $-2 + 3\sqrt{3} + \log_e \sqrt{3}$ - 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 $\int_{1/2}^0 \frac{-2}{(1-t)(1+t)^2} dt = \int_0^{1/2} \frac{2}{(1-t)(1+t)^2} dt$ , ensuring the boundary values are not reversed.
D · correct: $\frac{10}{3} - \sqrt{3} + \log_e \sqrt{3}$ - None. This is the correct evaluation. Splitting the integrand into $2\int_{\pi/3}^{\pi/2} \frac{dx}{\sin x(1+\cos x)} + 3\int_{\pi/3}^{\pi/2} \frac{dx}{1+\cos x}$ leads directly to $\frac{10}{3} - \sqrt{3} + \log_e \sqrt{3}$ .

Step-by-step solution

given: Definite integral $I = \int_{\pi/3}^{\pi/2} \frac{2 + 3 \sin x}{\sin x(1 + \cos x)} \, dx$ .
goal: Evaluate the exact value of $I$ .
approach: Split the integrand into two separate integrals: $I_1 = \int_{\pi/3}^{\pi/2} \frac{2}{\sin x(1 + \cos x)} \, dx$ and $I_2 = \int_{\pi/3}^{\pi/2} \frac{3}{1 + \cos x} \, dx$ . 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 $dx = \frac{2}{1+t^2} \, dt$ , $\sin x = \frac{2t}{1+t^2}$ , and $1 + \cos x = \frac{2}{1+t^2}$ . The limits transform from $x = \pi/3 \implies t = 1/\sqrt{3}$ to $x = \pi/2 \implies t = 1$ . The integral becomes: $I_1 = \int_{1/\sqrt{3}}^1 \frac{2}{\frac{2t}{1+t^2} \cdot \frac{2}{1+t^2}} \cdot \frac{2}{1+t^2} \, dt = \int_{1/\sqrt{3}}^1 \frac{1 + t^2}{t} \, dt = \int_{1/\sqrt{3}}^1 \left(\frac{1}{t} + t\right) dt$ Evaluating: $\left[\ln t + \frac{t^2}{2}\right]_{1/\sqrt{3}}^1 = \left(0 + \frac{1}{2}\right) - \left(\ln(1/\sqrt{3}) + \frac{1}{6}\right) = \frac{1}{3} + \ln\sqrt{3}$ .
execute: For $I_2$ , we have: $I_2 = 3 \int_{\pi/3}^{\pi/2} \frac{dx}{2\cos^2(x/2)} = \frac{3}{2} \int_{\pi/3}^{\pi/2} \sec^2(x/2) \, dx = 3 \left[\tan(x/2)\right]_{\pi/3}^{\pi/2} = 3 \left(1 - \frac{1}{\sqrt{3}}\right) = 3 - \sqrt{3}$ .
execute: Sum $I_1$ and $I_2$ to find $I$ : $I = I_1 + I_2 = \left(\frac{1}{3} + \ln\sqrt{3}\right) + (3 - \sqrt{3}) = \frac{10}{3} - \sqrt{3} + \log_e \sqrt{3}$ .
verify: Check at endpoints and order of magnitude: on $[\pi/3, \pi/2]$ , $\sin x \in [\sqrt{3}/2, 1]$ 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 $\pi/6 \approx 0.5236$ . 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.

Integral Calculus: JEE Main Mathematics Question with Solution

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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