The value of \int\limits_0^2\pi [ \sin 2x( 1 + \cos 3x ) ] dx , where [t] denotes the greatest integer function is :

Definite Integration PYQ — JEE Main 2019 · 10 Apr · Shift 1 · Q40 with Solution

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Correct answer: D

Let $I=\int_0^{2\pi} [\sin 2x(1+\cos 3x)]\,dx.$ We simplify the expression inside the greatest integer function.
Use $1+\cos 3x=2\cos^2\frac{3x}{2}, \qquad \sin 2x=2\sin x\cos x.$ A more useful identity here is: $\sin 2x(1+\cos 3x)=\sin 2x+\sin 2x\cos 3x.$ Now, $\sin 2x\cos 3x=\frac12[\sin(5x)+\sin(-x)]=\frac12(\sin 5x-\sin x).$ So $\sin 2x(1+\cos 3x)=\sin 2x+\frac12\sin 5x-\frac12\sin x.$ This is not yet enough directly for GIF evaluation.
Instead, use the product-to-sum identity in a better way: $\sin 2x(1+\cos 3x)=2\sin 2x\cos^2\frac{3x}{2}.$ But the cleanest simplification comes from writing $1+\cos 3x=2\cos^2\frac{3x}{2}$ and checking the range of the product. Since $-1\le \sin 2x\le 1, \qquad 0\le 1+\cos 3x\le 2,$ we get $-2\le \sin 2x(1+\cos 3x)\le 2.$ Hence the greatest integer can only be one of $-2,-1,0,1.$
Now observe an important symmetry. Define $f(x)=\sin 2x(1+\cos 3x).$ Then $f(2\pi-x)=\sin(4\pi-2x)\bigl(1+\cos(6\pi-3x)\bigr)=-\sin 2x(1+\cos 3x)=-f(x).$ So the function is antisymmetric about $x=\pi$ .

Thus, for each pair $x$ and $2\pi-x$ , the values are $t$ and $-t$ . Therefore

-1, & t\notin \mathbb Z,\\ 0, & t\in \mathbb Z. \end{cases}$$ Since $f(x)$ is continuous and takes integer values only at isolated points (which have measure zero), we have almost everywhere $$[f(x)]+[f(2\pi-x)]=-1.$$ 5. Therefore, $$2I=\int_0^{2\pi} [f(x)]\,dx+\int_0^{2\pi}[f(2\pi-x)]\,dx.$$ By substitution $u=2\pi-x$, the second integral equals $$\int_0^{2\pi}[f(u)]\,du=I.$$ So $$2I=\int_0^{2\pi}\big([f(x)]+[-f(x)]\big)\,dx.$$ Almost everywhere the integrand is $-1$, hence $$2I=\int_0^{2\pi}(-1)\,dx=-2\pi.$$ Therefore, $$I=-\pi.$$ 6. So the correct option is $$\boxed{\text{D }(-\pi)}.$$

View written solutionFree

Correct answer: D

Let $I=\int_0^{2\pi} [\sin 2x(1+\cos 3x)]\,dx.$ We simplify the expression inside the greatest integer function.
Use $1+\cos 3x=2\cos^2\frac{3x}{2}, \qquad \sin 2x=2\sin x\cos x.$ A more useful identity here is: $\sin 2x(1+\cos 3x)=\sin 2x+\sin 2x\cos 3x.$ Now, $\sin 2x\cos 3x=\frac12[\sin(5x)+\sin(-x)]=\frac12(\sin 5x-\sin x).$ So $\sin 2x(1+\cos 3x)=\sin 2x+\frac12\sin 5x-\frac12\sin x.$ This is not yet enough directly for GIF evaluation.
Instead, use the product-to-sum identity in a better way: $\sin 2x(1+\cos 3x)=2\sin 2x\cos^2\frac{3x}{2}.$ But the cleanest simplification comes from writing $1+\cos 3x=2\cos^2\frac{3x}{2}$ and checking the range of the product. Since $-1\le \sin 2x\le 1, \qquad 0\le 1+\cos 3x\le 2,$ we get $-2\le \sin 2x(1+\cos 3x)\le 2.$ Hence the greatest integer can only be one of $-2,-1,0,1.$
Now observe an important symmetry. Define $f(x)=\sin 2x(1+\cos 3x).$ Then $f(2\pi-x)=\sin(4\pi-2x)\bigl(1+\cos(6\pi-3x)\bigr)=-\sin 2x(1+\cos 3x)=-f(x).$ So the function is antisymmetric about $x=\pi$ .

Thus, for each pair $x$ and $2\pi-x$ , the values are $t$ and $-t$ . Therefore

-1, & t\notin \mathbb Z,\\ 0, & t\in \mathbb Z. \end{cases}$$ Since $f(x)$ is continuous and takes integer values only at isolated points (which have measure zero), we have almost everywhere $$[f(x)]+[f(2\pi-x)]=-1.$$ 5. Therefore, $$2I=\int_0^{2\pi} [f(x)]\,dx+\int_0^{2\pi}[f(2\pi-x)]\,dx.$$ By substitution $u=2\pi-x$, the second integral equals $$\int_0^{2\pi}[f(u)]\,du=I.$$ So $$2I=\int_0^{2\pi}\big([f(x)]+[-f(x)]\big)\,dx.$$ Almost everywhere the integrand is $-1$, hence $$2I=\int_0^{2\pi}(-1)\,dx=-2\pi.$$ Therefore, $$I=-\pi.$$ 6. So the correct option is $$\boxed{\text{D }(-\pi)}.$$