Visual SolutionPYQ 2024 · Jan Shift 2Standard

Question

The value of the integral

\int_0^{\pi/4} \frac{x \, dx}{\sin^4(2x) + \cos^4(2x)}

equals:

(A)

\frac{\sqrt{2}\pi^2}{8}

(B)

\frac{\sqrt{2}\pi^2}{16}

(C)

\frac{\sqrt{2}\pi^2}{32}

(D)

\frac{\sqrt{2}\pi^2}{64}

Solution Path

Substitution

t=2x

, then King's property.

I = \frac{\pi}{8} \cdot \frac{\pi}{\sqrt{2}} = \frac{\sqrt{2}\,\pi^2}{32}

01Question Setup

1/4

Evaluate the given definite integral involving

\sin^4

and

\cos^4

terms.

I = \;?

02Substitution

2/4

Let

t = 2x

. Transform the integral. Use the identity

\sin^4 t + \cos^4 t = 1 - \frac{\sin^2 2t}{2}

\sin^4 t + \cos^4 t = 1 - \frac{\sin^2 2t}{2}

03King's PropertyKEY INSIGHT

3/4

Apply King's property to eliminate

t

. Evaluate the remaining integral as

\frac{\pi}{\sqrt{2}}

. Result:

\frac{\sqrt{2}\,\pi^2}{32}

I = \frac{\sqrt{2}\,\pi^2}{32}

04Final Answer

4/4

King's property simplifies the integral elegantly.

\boxed{\dfrac{\sqrt{2}\,\pi^2}{32}}

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

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

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