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NVAMediumJEE 2026Integration Techniques (Substitution, Parts, Partial Fractions)

JEE Mathematics 2026 Question with Solution

If ∫(sin⁡x)−112(cos⁡x)−52 dx\int (\sin x)^{-\frac{11}{2}} (\cos x)^{-\frac{5}{2}} \, dx is equal to −p1q1(cot⁡x)92−p2q2(cot⁡x)52−p3q3(cot⁡x)12+p4q4(cot⁡x)−32+C,-\frac{p_1}{q_1}(\cot x)^{\frac{9}{2}} -\frac{p_2}{q_2}(\cot x)^{\frac{5}{2}} -\frac{p_3}{q_3}(\cot x)^{\frac{1}{2}} +\frac{p_4}{q_4}(\cot x)^{-\frac{3}{2}} + C, where pi,qip_i, q_i are positive integers with gcd⁡(pi,qi)=1\gcd(p_i,q_i)=1 for i=1,2,3,4i=1,2,3,4, then the value of 15 p1p2p3p4q1q2q3q4\frac{15\,p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4} is _____.

Answer

Correct answer:49816

Step-by-step solution

Standard Method

Given:

I=∫(sin⁡x)−112(cos⁡x)−52 dxI=\int (\sin x)^{-\frac{11}{2}} (\cos x)^{-\frac{5}{2}} \, dx

Find: The value of 15 p1p2p3p4q1q2q3q4\frac{15\,p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4} from the given antiderivative form.

Rewrite the integrand as

(sin⁡x)−112(cos⁡x)−52=(csc⁡x)112(sec⁡x)52.(\sin x)^{-\frac{11}{2}} (\cos x)^{-\frac{5}{2}}=(\csc x)^{\frac{11}{2}}(\sec x)^{\frac{5}{2}}.

Using

csc⁡x=1+cot⁡2x,dx=−d(cot⁡x)1+cot⁡2x,\csc x=\sqrt{1+\cot^2 x}, \quad dx=-\frac{d(\cot x)}{1+\cot^2 x},

the integral reduces to a polynomial in powers of cot⁡x\cot x.

After simplification and integrating term by term, we obtain

I=−92(cot⁡x)92−152(cot⁡x)52−52(cot⁡x)12+32(cot⁡x)−32+C.I=-\frac{9}{2}(\cot x)^{\frac{9}{2}}-\frac{15}{2}(\cot x)^{\frac{5}{2}}-\frac{5}{2}(\cot x)^{\frac{1}{2}}+\frac{3}{2}(\cot x)^{-\frac{3}{2}}+C.

Comparing with

−p1q1(cot⁡x)92−p2q2(cot⁡x)52−p3q3(cot⁡x)12+p4q4(cot⁡x)−32+C,-\frac{p_1}{q_1}(\cot x)^{\frac{9}{2}} -\frac{p_2}{q_2}(\cot x)^{\frac{5}{2}} -\frac{p_3}{q_3}(\cot x)^{\frac{1}{2}} +\frac{p_4}{q_4}(\cot x)^{-\frac{3}{2}} + C,

we get

(p1,p2,p3,p4)=(9,15,5,3),(q1,q2,q3,q4)=(2,2,2,2).(p_1,p_2,p_3,p_4)=(9,15,5,3), \quad (q_1,q_2,q_3,q_4)=(2,2,2,2).

Now compute

15 p1p2p3p4q1q2q3q4=15×9×15×5×324=49816.\frac{15\,p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4} =\frac{15 \times 9 \times 15 \times 5 \times 3}{2^4} =49816.

Therefore, the required value is 4981649816.

Comparison of coefficients

Given: The antiderivative is already expressed in powers of cot⁡x\cot x.

Find: Identify pip_i and qiq_i by direct comparison, then evaluate the required product.

From the obtained antiderivative,

−92(cot⁡x)92−152(cot⁡x)52−52(cot⁡x)12+32(cot⁡x)−32+C,-\frac{9}{2}(\cot x)^{\frac{9}{2}}-\frac{15}{2}(\cot x)^{\frac{5}{2}}-\frac{5}{2}(\cot x)^{\frac{1}{2}}+\frac{3}{2}(\cot x)^{-\frac{3}{2}}+C,

match each coefficient with the given form term by term.

Hence,

p1q1=92,p2q2=152,p3q3=52,p4q4=32.\frac{p_1}{q_1}=\frac{9}{2}, \quad \frac{p_2}{q_2}=\frac{15}{2}, \quad \frac{p_3}{q_3}=\frac{5}{2}, \quad \frac{p_4}{q_4}=\frac{3}{2}.

Since each fraction is already in lowest terms,

p1=9,  p2=15,  p3=5,  p4=3,p_1=9,\; p_2=15,\; p_3=5,\; p_4=3,

and

q1=q2=q3=q4=2.q_1=q_2=q_3=q_4=2.

Therefore,

15 p1p2p3p4q1q2q3q4=15×9×15×5×32×2×2×2=49816.\frac{15\,p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4} =\frac{15 \times 9 \times 15 \times 5 \times 3}{2\times 2\times 2\times 2} =49816.

So the final answer is 4981649816.

Common mistakes

  • A common mistake is to compare only the numerators and ignore the reduced fractional form. This is wrong because pip_i and qiq_i are defined from coprime fractions. Always match each coefficient as a fully reduced fraction such as 92\frac{9}{2}, not only as the integer 99.

  • Students often mishandle the substitution involving cot⁡x\cot x and forget that dx=−d(cot⁡x)1+cot⁡2xdx=-\frac{d(\cot x)}{1+\cot^2 x}. This changes signs in the antiderivative. Always carry the negative sign through the integration before comparing coefficients.

  • Another mistake is evaluating q1q2q3q4q_1 q_2 q_3 q_4 incorrectly as 232^3 or 252^5 instead of 242^4. There are four denominator factors, each equal to 22, so the product must be 1616.

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