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Definite Integration question

2024 · 27 Jan · Shift 2 · Q49

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Definite Integration
2024 · 27 Jan · Shift 2 · Q49

Definite Integration question

2024 · 27 Jan · Shift 2 · Q49

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1

For

0 \lt \mathrm{a} \lt 1

, the value of the integral

\int\limits_0^\pi \frac{\mathrm{d} x}{1-2 \mathrm{a} \cos x+\mathrm{a}^2}

is :

A
$\frac{\pi^2}{\pi+a^2}$
B
$\frac{\pi^2}{\pi-a^2}$
C
$\frac{\pi}{1-\mathrm{a}^2}$
D
$\frac{\pi}{1+\mathrm{a}^2}$

View written solutionFree

Correct answer: C

We need to evaluate $I=\int_0^\pi \frac{dx}{1-2a\cos x+a^2}, \qquad 0<a<1.$
Rewrite the denominator in the standard form: $1-2a\cos x+a^2=(1-a)^2+2a(1-\cos x).$ But the most useful standard result is $\int_0^\pi \frac{dx}{A+B\cos x}=\frac{\pi}{\sqrt{A^2-B^2}} \quad (A>|B|).$

Here, $1-2a\cos x+a^2=(1+a^2)-2a\cos x.$ So we identify $A=1+a^2, \qquad B=-2a.$

Check the condition: $A>|B| \iff 1+a^2>2a \iff (1-a)^2>0,$ which is true since $0<a<1$ .
Apply the formula: $I=\frac{\pi}{\sqrt{(1+a^2)^2-(-2a)^2}}.$ Now simplify: $(1+a^2)^2-4a^2 = 1+2a^2+a^4-4a^2 = 1-2a^2+a^4=(1-a^2)^2.$ Hence, $I=\frac{\pi}{\sqrt{(1-a^2)^2}}=\frac{\pi}{1-a^2},$ because $0<a<1 \implies 1-a^2>0$ .
Therefore, $\int_0^\pi \frac{dx}{1-2a\cos x+a^2}=\frac{\pi}{1-a^2}.$
Option check:

A: $\dfrac{\pi^2}{\pi+a^2}$ — incorrect
B: $\dfrac{\pi^2}{\pi-a^2}$ — incorrect
C: $\dfrac{\pi}{1-a^2}$ — correct
D: $\dfrac{\pi}{1+a^2}$ — incorrect

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