Why does the boundary term cancel out?

The evaluation of yields , which cancels the term added in the numerator.

JEE MainMathematics

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Integral Calculus: JEE Main Mathematics Question with Solution

The value of

\frac{\mathrm{e}^{-\frac{\pi}{4}} + \int_0^{\frac{\pi}{4}} \mathrm{e}^{-x}\tan^{50}x\,dx}{\int_0^{\frac{\pi}{4}} \mathrm{e}^{-x}\left(\tan^{49}x + \tan^{51}x\right)dx}

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A50B49C51D25

Step-by-step solutionView

Correct answer

Using integration by parts on $\int_0^{\pi/4} \mathrm{e}^{-x}\tan^{50}x\,dx$ simplifies the numerator to $50$ times the denominator, giving an overall value of $50$ .

Option analysis

Why each option works or fails

A · 50

None. This is the correct value. Integrating $\mathrm{e}^{-x}$ while differentiating $\tan^{50}x$ brings down a factor of $50$ , leaving the boundary term and the integral precisely proportional to the denominator.

B · 49

Believing that differentiating $\tan^n x$ brings down an exponent of $(n-1)$ instead of $n$ , or incorrectly factoring out $\tan^{49}x$ . Recall that $\frac{d}{dx}(\tan^n x) = n\tan^{n-1}x\sec^2 x$ . For $n=50$ , the coefficient generated is $50$ , not $49$ .

C · 51

Confusing differentiation of power functions with integration, mistakenly using $(n+1) = 51$ as the coefficient. Integration by parts differentiates $\tan^{50}x$ , which multiplies the term by $50$ , not $50+1=51$ .

D · 25

Assuming an extra factor of $2$ is introduced when writing $\tan^{49}x + \tan^{51}x = \tan^{49}x\sec^2 x$ , dividing $50$ by $2$ . Use the identity $1 + \tan^2 x = \sec^2 x$ directly without dividing by $2$ : $\tan^{49}x(1 + \tan^2 x) = \tan^{49}x\sec^2 x$ .

Reviewed route

Solution

StepWorking

01given

Expression: $E = \frac{\mathrm{e}^{-\pi/4} + \int_0^{\pi/4} \mathrm{e}^{-x}\tan^{50}x\,dx}{\int_0^{\pi/4} \mathrm{e}^{-x}\left(\tan^{49}x + \tan^{51}x\right)dx}$ .

02goal

Evaluate the value of the rational expression $E$ .

03approach

Apply integration by parts to $I_{50} = \int_0^{\pi/4} \mathrm{e}^{-x}\tan^{50}x\,dx$ with $u = \tan^{50}x$ and $dv = \mathrm{e}^{-x}dx$ . The derivative creates $\sec^2 x = 1 + \tan^2 x$ , naturally producing the denominator term $(\tan^{49}x + \tan^{51}x)$ .

04execute

Integrate by parts: $\int_0^{\pi/4} \mathrm{e}^{-x} \tan^{50}x\,dx = \left[ -\mathrm{e}^{-x}\tan^{50}x \right]_0^{\pi/4} - \int_0^{\pi/4} (-\mathrm{e}^{-x}) \cdot 50\tan^{49}x \sec^2x\,dx$ $= -\mathrm{e}^{-\pi/4}(1)^{50} - 0 + 50 \int_0^{\pi/4} \mathrm{e}^{-x}\tan^{49}x(1 + \tan^2x)\,dx$ $= -\mathrm{e}^{-\pi/4} + 50 \int_0^{\pi/4} \mathrm{e}^{-x}(\tan^{49}x + \tan^{51}x)\,dx$

05execute

Rearranging the equation: $\mathrm{e}^{-\pi/4} + \int_0^{\pi/4} \mathrm{e}^{-x}\tan^{50}x\,dx = 50 \int_0^{\pi/4} \mathrm{e}^{-x}(\tan^{49}x + \tan^{51}x)\,dx$ Dividing both sides by the denominator integral yields $E = 50$ .

✓verify

For a general power $n$ , $\int_0^{\pi/4} \mathrm{e}^{-x}\tan^n x\,dx = -\mathrm{e}^{-\pi/4} + n \int_0^{\pi/4} \mathrm{e}^{-x}(\tan^{n-1}x + \tan^{n+1}x)\,dx$ , so the ratio is identically $n = 50$ .

✓ 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: 7 September 2026

Quick checks

Students also ask

Why does the boundary term

\mathrm{e}^{-\pi/4}

cancel out?

The evaluation of $[-\mathrm{e}^{-x}\tan^{50}x]_0^{\pi/4}$ yields $-\mathrm{e}^{-\pi/4}$ , which cancels the $+\mathrm{e}^{-\pi/4}$ term added in the numerator.

Answer

Using integration by parts on $\int_0^{\pi/4} \mathrm{e}^{-x}\tan^{50}x\,dx$ simplifies the numerator to $50$ times the denominator, giving an overall value of $50$ .

Why each option works or fails

A · correct: 50 - None. This is the correct value. Integrating $\mathrm{e}^{-x}$ while differentiating $\tan^{50}x$ brings down a factor of $50$ , leaving the boundary term and the integral precisely proportional to the denominator.
B: 49 - Believing that differentiating $\tan^n x$ brings down an exponent of $(n-1)$ instead of $n$ , or incorrectly factoring out $\tan^{49}x$ . Recall that $\frac{d}{dx}(\tan^n x) = n\tan^{n-1}x\sec^2 x$ . For $n=50$ , the coefficient generated is $50$ , not $49$ .
C: 51 - Confusing differentiation of power functions with integration, mistakenly using $(n+1) = 51$ as the coefficient. Integration by parts differentiates $\tan^{50}x$ , which multiplies the term by $50$ , not $50+1=51$ .
D: 25 - Assuming an extra factor of $2$ is introduced when writing $\tan^{49}x + \tan^{51}x = \tan^{49}x\sec^2 x$ , dividing $50$ by $2$ . Use the identity $1 + \tan^2 x = \sec^2 x$ directly without dividing by $2$ : $\tan^{49}x(1 + \tan^2 x) = \tan^{49}x\sec^2 x$ .

Step-by-step solution

given: Expression: $E = \frac{\mathrm{e}^{-\pi/4} + \int_0^{\pi/4} \mathrm{e}^{-x}\tan^{50}x\,dx}{\int_0^{\pi/4} \mathrm{e}^{-x}\left(\tan^{49}x + \tan^{51}x\right)dx}$ .
goal: Evaluate the value of the rational expression $E$ .
approach: Apply integration by parts to $I_{50} = \int_0^{\pi/4} \mathrm{e}^{-x}\tan^{50}x\,dx$ with $u = \tan^{50}x$ and $dv = \mathrm{e}^{-x}dx$ . The derivative creates $\sec^2 x = 1 + \tan^2 x$ , naturally producing the denominator term $(\tan^{49}x + \tan^{51}x)$ .
execute: Integrate by parts: $\int_0^{\pi/4} \mathrm{e}^{-x} \tan^{50}x\,dx = \left[ -\mathrm{e}^{-x}\tan^{50}x \right]_0^{\pi/4} - \int_0^{\pi/4} (-\mathrm{e}^{-x}) \cdot 50\tan^{49}x \sec^2x\,dx$ $= -\mathrm{e}^{-\pi/4}(1)^{50} - 0 + 50 \int_0^{\pi/4} \mathrm{e}^{-x}\tan^{49}x(1 + \tan^2x)\,dx$ $= -\mathrm{e}^{-\pi/4} + 50 \int_0^{\pi/4} \mathrm{e}^{-x}(\tan^{49}x + \tan^{51}x)\,dx$
execute: Rearranging the equation: $\mathrm{e}^{-\pi/4} + \int_0^{\pi/4} \mathrm{e}^{-x}\tan^{50}x\,dx = 50 \int_0^{\pi/4} \mathrm{e}^{-x}(\tan^{49}x + \tan^{51}x)\,dx$ Dividing both sides by the denominator integral yields $E = 50$ .
verify: For a general power $n$ , $\int_0^{\pi/4} \mathrm{e}^{-x}\tan^n x\,dx = -\mathrm{e}^{-\pi/4} + n \int_0^{\pi/4} \mathrm{e}^{-x}(\tan^{n-1}x + \tan^{n+1}x)\,dx$ , so the ratio is identically $n = 50$ .

Integral Calculus: JEE Main Mathematics Question with Solution

Using integration by parts on ∫0π/4e−xtan⁡50x dx\int_0^{\pi/4} \mathrm{e}^{-x}\tan^{50}x\,dx∫0π/4​e−xtan50xdx simplifies the numerator to 505050 times the denominator, giving an overall value of 505050.

01given

02goal

03approach

04execute

05execute

✓verify

Students also ask

Using integration by parts on $\int_0^{\pi/4} \mathrm{e}^{-x}\tan^{50}x\,dx$ simplifies the numerator to $50$ times the denominator, giving an overall value of $50$ .