JEE MainMathematics

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Definite Integral Involving Integration by Parts and Trigonometric Identity

Q: Why does the boundary term e^(-π/4) cancel out?

The evaluation of [-e^(-x)tan^50x]_0^(π/4) yields -e^(-π/4), which cancels the +e^(-π/4) term added in the numerator.

The value of (e^(-π/4) + integral from 0 to π/4 of e^(-x)tan^50x dx)/(integral from 0 to π/4 of e^(-x)(tan^49x + tan^51x)dx) is

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A50B49C51D25

Step-by-step solutionView

Correct answer

Using integration by parts on $\int_{0}^{π / 4} e^{- x} \tan^{50} x d x$ simplifies the numerator to $50$ times the denominator, giving an overall value of $50$ .

Option analysis

Why each option works or fails

A · 50

This is the correct value. Integrating $e^{- x}$ 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 $\frac{d}{d x} (\tan^{n} x) = n \tan^{n - 1} x \sec^{2} x$ . 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: $E = \frac{e^{- π / 4} + \int_{0}^{π / 4} e^{- x} \tan^{50} x d x}{\int_{0}^{π / 4} e^{- x} (\tan^{49} x + \tan^{51} x) d x}$ .

02Goal

Evaluate the value of the rational expression $E$ .

03Approach

Apply integration by parts to $I_{50} = \int_{0}^{π / 4} e^{- x} \tan^{50} x d x$ with $u = \tan^{50} x$ and $d v = e^{- x} d x$ . 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: $\int_{0}^{π / 4} e^{- x} \tan^{50} x d x = {[- e^{- x} \tan^{50} x]}_{0}^{π / 4} - \int_{0}^{π / 4} (- e^{- x}) \cdot 50 \tan^{49} x \sec^{2} x d x$ $= - e^{- π / 4} (1)^{50} - 0 + 50 \int_{0}^{π / 4} e^{- x} \tan^{49} x (1 + \tan^{2} x) d x$ $= - e^{- π / 4} + 50 \int_{0}^{π / 4} e^{- x} (\tan^{49} x + \tan^{51} x) d x$

05Execute

Rearranging the equation: $e^{- π / 4} + \int_{0}^{π / 4} e^{- x} \tan^{50} x d x = 50 \int_{0}^{π / 4} e^{- x} (\tan^{49} x + \tan^{51} x) d x$ Dividing both sides by the denominator integral yields $E = 50$ .

✓Verify

For a general power $n$ , $\int_{0}^{π / 4} e^{- x} \tan^{n} x d x = - e^{- π / 4} + n \int_{0}^{π / 4} e^{- x} (\tan^{n - 1} x + \tan^{n + 1} x) d x$ , so the ratio is identically $n = 50$ .

Hints that build this answer step by step

How can the integrand in the denominator, $\tan^{49} x + \tan^{51} x$ , be simplified using a standard trigonometric identity?
$\tan^{49} x (1 + \tan^{2} x) = \tan^{49} x \sec^{2} x$
Applying integration by parts to $I = \int_{0}^{π / 4} e^{- x} \tan^{50} x d x$ with $u = \tan^{50} x$ and $d v = e^{- x} d x$ , what is the resulting expression?
$I = - e^{- π / 4} + 50 \int_{0}^{π / 4} e^{- x} \tan^{49} x \sec^{2} x d x$
Substitute the expression for $I$ into the numerator $e^{- π / 4} + I$ and divide by the denominator. What is the final value?
50

✓ 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: 7 September 2026

Quick checks

Students also ask

Why does the boundary term

e^{- π / 4}

cancel out?

The evaluation of $[- e^{- x} \tan^{50} x]_{0}^{π / 4}$ yields $- e^{- π / 4}$ , which cancels the $+ e^{- π / 4}$ term added in the numerator.

Answer

Using integration by parts on $\int_{0}^{π / 4} e^{- x} \tan^{50} x d x$ simplifies the numerator to $50$ times the denominator, giving an overall value of $50$ .

Why each option works or fails

A · correct: 50 - This is the correct value. Integrating $e^{- x}$ 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 $\frac{d}{d x} (\tan^{n} x) = n \tan^{n - 1} x \sec^{2} x$ . 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: $E = \frac{e^{- π / 4} + \int_{0}^{π / 4} e^{- x} \tan^{50} x d x}{\int_{0}^{π / 4} e^{- x} (\tan^{49} x + \tan^{51} x) d x}$ .
Goal: Evaluate the value of the rational expression $E$ .
Approach: Apply integration by parts to $I_{50} = \int_{0}^{π / 4} e^{- x} \tan^{50} x d x$ with $u = \tan^{50} x$ and $d v = e^{- x} d x$ . 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: $\int_{0}^{π / 4} e^{- x} \tan^{50} x d x = {[- e^{- x} \tan^{50} x]}_{0}^{π / 4} - \int_{0}^{π / 4} (- e^{- x}) \cdot 50 \tan^{49} x \sec^{2} x d x$ $= - e^{- π / 4} (1)^{50} - 0 + 50 \int_{0}^{π / 4} e^{- x} \tan^{49} x (1 + \tan^{2} x) d x$ $= - e^{- π / 4} + 50 \int_{0}^{π / 4} e^{- x} (\tan^{49} x + \tan^{51} x) d x$
Execute: Rearranging the equation: $e^{- π / 4} + \int_{0}^{π / 4} e^{- x} \tan^{50} x d x = 50 \int_{0}^{π / 4} e^{- x} (\tan^{49} x + \tan^{51} x) d x$ Dividing both sides by the denominator integral yields $E = 50$ .
Verify: For a general power $n$ , $\int_{0}^{π / 4} e^{- x} \tan^{n} x d x = - e^{- π / 4} + n \int_{0}^{π / 4} e^{- x} (\tan^{n - 1} x + \tan^{n + 1} x) d x$ , so the ratio is identically $n = 50$ .

Definite Integral Involving Integration by Parts and Trigonometric Identity

What feels right?

Using integration by parts on $\int_{0}^{π / 4} e^{- x} \tan^{50} x d x$ simplifies the numerator to $50$ times the denominator, giving an overall value of $50$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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