Visual SolutionPYQ 2024 · Apr Shift 1Standard

Question

The integral

\int_0^{\pi/4} \frac{\cos^2 x \sin^2 x}{(\cos^3 x + \sin^3 x)^2} \, dx

is equal to:

(A)

\frac{1}{6}

(B)

\frac{1}{3}

(C)

\frac{1}{12}

(D)

\frac{1}{9}

Solution Path

Divide by

\cos^6 x

to get tan form. Substitute

t = 1 + \tan^3 x

. Integral

= \frac{1}{3}\int_1^2 t^{-2}\,dt = 1/6

01Question Setup

1/4

Evaluate

\int_0^{\pi/4} \dfrac{\cos^2 x \sin^2 x}{(\cos^3 x + \sin^3 x)^2} \, dx

Evaluate the definite integral

02Rewrite in tan

2/4

Divide numerator and denominator by

\cos^6 x

. The integrand becomes

\dfrac{\tan^2 x \cdot \sec^2 x}{(1 + \tan^3 x)^2}

= \dfrac{\tan^2 x \cdot \sec^2 x}{(1 + \tan^3 x)^2}

03SubstitutionKEY INSIGHT

3/4

Let

t = 1 + \tan^3 x

dt = 3\tan^2 x \sec^2 x \, dx

. Integral becomes

\tfrac{1}{3}\int_1^2 t^{-2} dt = \tfrac{1}{3} \cdot \tfrac{1}{2} = \tfrac{1}{6}

I = \dfrac{1}{3}\int_1^2 \dfrac{dt}{t^2} = \dfrac{1}{6}

04Final Answer

4/4

The substitution

t = 1 + \tan^3 x

collapses the integral to a simple power rule.

I = \boxed{\dfrac{1}{6}}

- Answer (A)

Concepts from this question1 concepts unlocked

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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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