Integral Calculus JEE 2014: Solving (2 Cosec x)^17

What was the (2 cosec x)^17 integral in JEE Advanced 2014?

JEE Advanced 2014 Paper 2 required evaluation of the definite integral from π/4 to π/2 of (2 cosec x)^17. It was tagged difficulty 4 (hard) with expected solving time of 300 seconds. The answer is $25740\ln (1+\sqrt{2})+\frac{16037316\sqrt{2}}{7}$.

This exact expression involves $\ln (1+\sqrt{2})$ and a rational multiple of $\sqrt{2}$. Students get stuck on the $e^u$ substitution for cosec x, the binomial expansion of the resulting even power, and the hyperbolic recurrence.

How do you solve the 2014 JEE Advanced integral of (2 cosec x)^17 using the official exponential substitution?

Set

Apply the substitution

At $x=\pi /4$, $u=\ln (1+\sqrt{2})$. At $x=\pi /2$, $u=0$. Let $L=\ln (1+\sqrt{2})$.

From the substitution derive $\csc x+\cot x=e^u$ and $\csc x-\cot x=e^{-u}$. Subtracting these gives $2\cot x=e^u-e^{-u}$.

Differentiate the substitution equation:

Solve for dx:

After substitution the factors simplify. The negative sign reverses the limits to yield

Expand using the binomial theorem:

The $k=8$ term is constant and integrates to $2\left(\genfrac{}{}{0ex}{}{16}{8}\right)L$. Since $\left(\genfrac{}{}{0ex}{}{16}{8}\right)=12870$ this logarithmic piece equals $25740L=25740\ln (1+\sqrt{2})$.

The remaining paired terms become

With $L=\ln (1+\sqrt{2})$ we have $\sinh L=1$ and $\cosh L=\sqrt{2}$, so $\sinh (2jL)=S_j\sqrt{2}$ where the sequence $S_j$ satisfies the recurrence $S_{j+1}=6S_j-S_{j-1}$ with $S_0=0$, $S_1=2$.

Computing the values:

• $S_1=2$
• $S_2=12$
• $S_3=70$
• $S_4=408$
• $S_5=2378$
• $S_6=13860$
• $S_7=80782$
• $S_8=470832$

After inserting each binomial coefficient, dividing by $j$, and summing, the algebraic coefficient evaluates to $\frac{16037316\sqrt{2}}{7}$.

Thus the exact value is

You can search every JEE Advanced paper from 2007 in the past-paper archive by chapter to locate this 2014 question with its worked solution.

What common mistake occurs in the substitution step for this integral?

One specific error occurs when students differentiate $e^u+e^{-u}=2\csc x$ but forget the $2\csc x$ factor present in the denominator of dx. They end up with an extra power of $(e^u+e^{-u})$ and obtain $I=\int (e^u+e^{-u}{)}^{17}du$ instead of the correct factor of 2 times the 16th power.

The resulting series carries different binomial coefficients and the summed rational term no longer simplifies to $16037316/7$. The final expression therefore fails to match the official answer even if the limits and hyperbolic pairing are applied later.

Why does the hyperbolic recurrence simplify the binomial sum so efficiently?

The recurrence originates directly from sinh addition formulas. Given $L=\ln (1+\sqrt{2})$, direct calculation shows $\sinh L=1$ and $\cosh L=\sqrt{2}$. Then $\sinh (2L)=2\sinh L\cosh L=2\sqrt{2}$, which sets $S_1=2$ when $\sinh (2jL)=S_j\sqrt{2}$.

The double-angle identity for $\sinh (2(j+1)L)$ expands using $\cosh (2L)=2{\cosh }^{2}L-1=3$. Collecting coefficients produces the clean linear recurrence $S_{j+1}=6S_j-S_{j-1}$ with initial values $S_0=0$, $S_1=2$.

The computed sequence is exactly:

• $S_1=2$
• $S_2=12$
• $S_3=70$
• $S_4=408$
• $S_5=2378$
• $S_6=13860$
• $S_7=80782$
• $S_8=470832$

Recurrence relations reduce computation time from O(n^2) to O(n) for the coefficients. After multiplying each $\left(\genfrac{}{}{0ex}{}{16}{8-j}\right)$ by $S_j/j$ and clearing the denominator, the total rational coefficient before $\sqrt{2}$ is $16037316/7$.

What are two related definite integral questions from the same chapter?

Evaluate ${\int }_{\pi /4}^{\pi /2}(2\csc x{)}^{5}dx$. The same exponential substitution converts the odd power directly into an even power of $(e^u+e^{-u})$ over the same limits from 0 to $L$.

A second question that demands binomial expansion after substitution and produces both logarithmic and rational terms is ${\int }_0^L(e^u+e^{-u}{)}^{8}du$ where $L=\ln (1+\sqrt{2})$. The central term again supplies a multiple of $L$ while the symmetric pairs collapse via the identical sinh recurrence.

For lower powers (n=5 or n=7) the recurrence method still works but becomes unnecessary overhead. Direct integration of the four or six surviving terms after expansion finishes faster. Solve both ways on mocks, then choose the recurrence route only when the exponent exceeds 9.

What are the key takeaways for JEE Advanced integral calculus from this 2014 question?

The substitution $e^u+e^{-u}=2\csc x$ converts odd powers of cosec x into even powers of $(e^u+e^{-u})$.

Symmetry in binomial coefficients allows all non-central terms to be written using sinh functions.

Recurrence relations reduce computation time from O(n^2) to O(n) for the coefficients.

The exact final form matches the official answer only when the factor of 2 and the sign of dx are handled correctly.

Apply the same substitution and S_j table to the next mock integral of this type. Photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one.

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