Visual SolutionPYQ 2024 · Apr Shift 1Standard

Question

The value of

\int_{-\pi}^{\pi} \frac{2y(1+\sin y)}{1+\cos^2 y} \, dy

is:

(A)

2\pi^2

(B)

\frac{\pi^2}{2}

(C)

\frac{\pi^2}{2}

(D)

\pi^2

Solution Path

Split into odd (vanishes) and even parts. King's property on

[0,\pi]

for the even part gives

\pi^2

. Total

= \pi^2

01Question Setup

1/4

Evaluate

\int_{-\pi}^{\pi} \frac{2y(1+\sin y)}{1+\cos^2 y}\,dy

\int_{-\pi}^{\pi} \frac{2y(1+\sin y)}{1+\cos^2 y}\,dy

02Split Odd/Even

2/4

Expand:

I_1 = \int \frac{2y}{1+\cos^2 y}\,dy

(odd function on symmetric interval, vanishes) and

I_2 = \int \frac{2y\sin y}{1+\cos^2 y}\,dy

(even function, doubles).

I_1 = 0 \;\text{(odd function)}

03King's PropertyKEY INSIGHT

3/4

Apply King's property on

[0, \pi]

for the even part: replace

y

with

\pi - y

. Add the two forms to eliminate

y

and simplify.

I_2 = \pi^2

04Final Answer

4/4

Total =

I_1 + I_2 = 0 + \pi^2 = \pi^2

\boxed{\pi^2}

Concepts from this question2 concepts unlocked

★

EASY

For definite integrals, replace x with (a+b-x) to exploit symmetry

\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx

If f(a+b-x) = -f(x), the integral is zero. If f(a+b-x) = f(x), the integral doubles

Symmetry integralsTrigonometric integralsPolynomial integrals

★

EASY

For symmetric limits, odd functions integrate to zero and even functions double

\int_{-a}^{a} f(x)\,dx = \begin{cases} 0 & f(-x) = -f(x) \\ 2\int_0^a f(x)\,dx & f(-x) = f(x) \end{cases}

Saves computation by detecting when an integral vanishes or can be simplified

Definite integralsTrigonometric integrals

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