Visual SolutionPYQ 2024 · Jan Shift 2Standard

Question

For

0 < a < 1

, the value of the integral

\int_0^{\pi} \frac{dx}{1 - 2a\cos x + a^2}

is:

(A)

\frac{\pi^2}{\pi + a^2}

(B)

\frac{\pi^2}{\pi - a^2}

(C)

\frac{\pi}{1-a^2}

(D)

\frac{\pi}{1+a^2}

Solution Path

Weierstrass sub

t = \tan\frac{x}{2}

reduces to standard arctan integral. Answer:

\frac{\pi}{1-a^2}

01Question Setup

1/4

For

0 < a < 1

, evaluate

\int_0^{\pi} \frac{dx}{1 - 2a\cos x + a^2}

I = \;?

02Weierstrass Sub

2/4

Let

t = \tan\frac{x}{2}

. Denominator becomes

\frac{(1-a)^2 + (1+a)^2 t^2}{1+t^2}

\int_0^{\infty} \frac{2\,dt}{(1-a)^2 + (1+a)^2 t^2}

03EvaluateKEY INSIGHT

3/4

Standard arctan integral:

\frac{\pi}{2A}

with

A = \frac{1-a}{1+a} \cdot (1-a)

. Result:

\frac{\pi}{1-a^2}

\frac{\pi}{1-a^2}

04Final Answer

4/4

\int_0^{\pi} \frac{dx}{1 - 2a\cos x + a^2} = \frac{\pi}{1-a^2}

\boxed{\frac{\pi}{1-a^2}}

Concepts from this question2 concepts unlocked

★

EASY

Replace a composite expression with a single variable to simplify the integrand

\int f(g(x))\,g'(x)\,dx = \int f(t)\,dt, \quad t = g(x)

Transforms complex integrands into standard forms that can be integrated directly

Trigonometric integralsRational functionsPower rule integrals

STANDARD

Decompose a rational function into simpler fractions before integrating

\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}

Converts complex rational integrands into a sum of elementary integrals

Rational integralsLogarithmic forms

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