\int\limits_0^\pi / 4 \frac\cos^2 x \sin^2 x(\cos^3 x+\sin^3 x)^2 dx is equal to

Definite Integration PYQ — JEE Main 2024 · 6 Apr · Shift 1 · Q44 with Solution

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Correct answer: B

Let $I=\int_0^{\pi/4} \frac{\cos^2 x\sin^2 x}{(\cos^3 x+\sin^3 x)^2}\,dx.$

We simplify the denominator using $\cos^3 x+\sin^3 x=(\cos x+\sin x)(\cos^2 x-\sin x\cos x+\sin^2 x).$ Since $\cos^2 x+\sin^2 x=1$ , this becomes $\cos^3 x+\sin^3 x=(\cos x+\sin x)(1-\sin x\cos x).$

So $I=\int_0^{\pi/4} \frac{\sin^2 x\cos^2 x}{(\sin x+\cos x)^2(1-\sin x\cos x)^2}\,dx.$

Use the substitution $t=\tan x, \qquad dx=\frac{dt}{1+t^2}.$ On the interval $x\in[0,\pi/4]$ , we have $t\in[0,1]$ .

Also, $\sin x=\frac{t}{\sqrt{1+t^2}},\qquad \cos x=\frac{1}{\sqrt{1+t^2}}.$ Hence $\sin x\cos x=\frac{t}{1+t^2},$ $\cos^3 x+\sin^3 x=\frac{1+t^3}{(1+t^2)^{3/2}}.$ Therefore,

=\frac{\frac{t^2}{(1+t^2)^2}}{\frac{(1+t^3)^2}{(1+t^2)^3}} =\frac{t^2(1+t^2)}{(1+t^3)^2}.$$ Multiplying by $dx=\frac{dt}{1+t^2}$, $$I=\int_0^1 \frac{t^2}{(1+t^3)^2}\,dt.$$ 3. Now substitute $$u=1+t^3 \quad\Rightarrow\quad du=3t^2\,dt,$$ so $$t^2\,dt=\frac{du}{3}.$$ When $t=0$, $u=1$; when $t=1$, $u=2$. Thus $$I=\frac13\int_1^2 u^{-2}\,du.$$ 4. Integrate: $$\int u^{-2}\,du=-u^{-1}.$$ So $$I=\frac13\left[-\frac1u\right]_1^2 =\frac13\left(-\frac12+1\right) =\frac13\cdot\frac12 =\frac16.$$ 5. Compare with options: - A: $\frac19$ - B: $\frac16$ ✅ - C: $\frac13$ - D: $\frac1{12}$ Therefore, the correct answer is $$\boxed{\frac16}.$$

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View written solutionFree

Correct answer: B

Let $I=\int_0^{\pi/4} \frac{\cos^2 x\sin^2 x}{(\cos^3 x+\sin^3 x)^2}\,dx.$

We simplify the denominator using $\cos^3 x+\sin^3 x=(\cos x+\sin x)(\cos^2 x-\sin x\cos x+\sin^2 x).$ Since $\cos^2 x+\sin^2 x=1$ , this becomes $\cos^3 x+\sin^3 x=(\cos x+\sin x)(1-\sin x\cos x).$

So $I=\int_0^{\pi/4} \frac{\sin^2 x\cos^2 x}{(\sin x+\cos x)^2(1-\sin x\cos x)^2}\,dx.$

Use the substitution $t=\tan x, \qquad dx=\frac{dt}{1+t^2}.$ On the interval $x\in[0,\pi/4]$ , we have $t\in[0,1]$ .

Also, $\sin x=\frac{t}{\sqrt{1+t^2}},\qquad \cos x=\frac{1}{\sqrt{1+t^2}}.$ Hence $\sin x\cos x=\frac{t}{1+t^2},$ $\cos^3 x+\sin^3 x=\frac{1+t^3}{(1+t^2)^{3/2}}.$ Therefore,

=\frac{\frac{t^2}{(1+t^2)^2}}{\frac{(1+t^3)^2}{(1+t^2)^3}} =\frac{t^2(1+t^2)}{(1+t^3)^2}.$$ Multiplying by $dx=\frac{dt}{1+t^2}$, $$I=\int_0^1 \frac{t^2}{(1+t^3)^2}\,dt.$$ 3. Now substitute $$u=1+t^3 \quad\Rightarrow\quad du=3t^2\,dt,$$ so $$t^2\,dt=\frac{du}{3}.$$ When $t=0$, $u=1$; when $t=1$, $u=2$. Thus $$I=\frac13\int_1^2 u^{-2}\,du.$$ 4. Integrate: $$\int u^{-2}\,du=-u^{-1}.$$ So $$I=\frac13\left[-\frac1u\right]_1^2 =\frac13\left(-\frac12+1\right) =\frac13\cdot\frac12 =\frac16.$$ 5. Compare with options: - A: $\frac19$ - B: $\frac16$ ✅ - C: $\frac13$ - D: $\frac1{12}$ Therefore, the correct answer is $$\boxed{\frac16}.$$

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