If \int\limits_0^\pi 5^\cos x(1 + \cos x\cos 3x + \cos ^2x + \cos ^3x\cos 3x)dx \over 1 + 5^\cos x = k\pi \over 16 , then k is equal to \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2023 · 1 Feb · Shift 2 · Q41 with Solution

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Correct answer: 13

Let $I=\int_0^\pi \frac{5^{\cos x}\left(1+\cos x\cos 3x+\cos^2x+\cos^3x\cos 3x\right)}{1+5^{\cos x}}\,dx.$

We simplify the trigonometric factor first.

Factor:

=(1+\cos^2x)(1+\cos x\cos 3x).$$ Now use $$\cos 3x=4\cos^3x-3\cos x,$$ so $$\cos x\cos 3x=4\cos^4x-3\cos^2x.$$ Hence $$1+\cos x\cos 3x=1-3\cos^2x+4\cos^4x=(1-\cos^2x)(1-4\cos^2x? )$$ But a better way is to use product-to-sum: $$\cos x\cos 3x=\frac{\cos 4x+\cos 2x}{2}.$$ Still, the key is symmetry of the integral, so let $$f(x)=1+\cos x\cos 3x+\cos^2x+\cos^3x\cos 3x.$$ 3. Use the standard substitution $x\mapsto \pi-x$. Since $$\cos(\pi-x)=-\cos x,\qquad 5^{\cos(\pi-x)}=5^{-\cos x},$$ we get $$\frac{5^{\cos(\pi-x)}}{1+5^{\cos(\pi-x)}}=\frac{5^{-\cos x}}{1+5^{-\cos x}}=\frac{1}{1+5^{\cos x}}.$$ Also, $$\cos 3(\pi-x)=\cos(3\pi-3x)=-\cos 3x.$$ Therefore \begin{align*} f(\pi-x)&=1+(-\cos x)(-\cos 3x)+\cos^2x+(-\cos x)^3(-\cos 3x)\\ &=1+\cos x\cos 3x+\cos^2x+\cos^3x\cos 3x\\ &=f(x). \end{align*} So $f(x)$ is unchanged. Thus, $$I=\int_0^\pi \frac{f(x)}{1+5^{\cos x}}\,dx.$$ Adding this with the original form, $$2I=\int_0^\pi f(x)\,dx,$$ because $$\frac{5^{\cos x}}{1+5^{\cos x}}+\frac{1}{1+5^{\cos x}}=1.$$ Hence $$I=\frac12\int_0^\pi \left(1+\cos x\cos 3x+\cos^2x+\cos^3x\cos 3x\right)\,dx.$$ 4. Now evaluate each term: - $$\int_0^\pi 1\,dx=\pi.$$ - $$\int_0^\pi \cos x\cos 3x\,dx=0,$$ by orthogonality. - $$\int_0^\pi \cos^2x\,dx=\frac\pi2.$$ - For $$\int_0^\pi \cos^3x\cos 3x\,dx,$$ use $$\cos 3x=4\cos^3x-3\cos x,$$ so $$\cos^3x\cos 3x=4\cos^6x-3\cos^4x.$$ Thus \begin{align*} \int_0^\pi \cos^3x\cos 3x\,dx &=4\int_0^\pi \cos^6x\,dx-3\int_0^\pi \cos^4x\,dx. \end{align*} Now the standard values are $$\int_0^\pi \cos^4x\,dx=\frac{3\pi}{8}, \qquad \int_0^\pi \cos^6x\,dx=\frac{5\pi}{16}.$$ Therefore \begin{align*} \int_0^\pi \cos^3x\cos 3x\,dx &=4\cdot \frac{5\pi}{16}-3\cdot \frac{3\pi}{8}\\ &=\frac{5\pi}{4}-\frac{9\pi}{8}\\ &=\frac{\pi}{8}. \end{align*} 5. So, \begin{align*} \int_0^\pi f(x)\,dx &=\pi+0+\frac\pi2+\frac\pi8\\ &=\frac{13\pi}{8}. \end{align*} Hence $$I=\frac12\cdot \frac{13\pi}{8}=\frac{13\pi}{16}.$$ 6. Comparing with $$I=\frac{k\pi}{16},$$ we get $$k=13.$$

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View written solutionFree

Correct answer: 13

Let $I=\int_0^\pi \frac{5^{\cos x}\left(1+\cos x\cos 3x+\cos^2x+\cos^3x\cos 3x\right)}{1+5^{\cos x}}\,dx.$

We simplify the trigonometric factor first.

Factor:

=(1+\cos^2x)(1+\cos x\cos 3x).$$ Now use $$\cos 3x=4\cos^3x-3\cos x,$$ so $$\cos x\cos 3x=4\cos^4x-3\cos^2x.$$ Hence $$1+\cos x\cos 3x=1-3\cos^2x+4\cos^4x=(1-\cos^2x)(1-4\cos^2x? )$$ But a better way is to use product-to-sum: $$\cos x\cos 3x=\frac{\cos 4x+\cos 2x}{2}.$$ Still, the key is symmetry of the integral, so let $$f(x)=1+\cos x\cos 3x+\cos^2x+\cos^3x\cos 3x.$$ 3. Use the standard substitution $x\mapsto \pi-x$. Since $$\cos(\pi-x)=-\cos x,\qquad 5^{\cos(\pi-x)}=5^{-\cos x},$$ we get $$\frac{5^{\cos(\pi-x)}}{1+5^{\cos(\pi-x)}}=\frac{5^{-\cos x}}{1+5^{-\cos x}}=\frac{1}{1+5^{\cos x}}.$$ Also, $$\cos 3(\pi-x)=\cos(3\pi-3x)=-\cos 3x.$$ Therefore \begin{align*} f(\pi-x)&=1+(-\cos x)(-\cos 3x)+\cos^2x+(-\cos x)^3(-\cos 3x)\\ &=1+\cos x\cos 3x+\cos^2x+\cos^3x\cos 3x\\ &=f(x). \end{align*} So $f(x)$ is unchanged. Thus, $$I=\int_0^\pi \frac{f(x)}{1+5^{\cos x}}\,dx.$$ Adding this with the original form, $$2I=\int_0^\pi f(x)\,dx,$$ because $$\frac{5^{\cos x}}{1+5^{\cos x}}+\frac{1}{1+5^{\cos x}}=1.$$ Hence $$I=\frac12\int_0^\pi \left(1+\cos x\cos 3x+\cos^2x+\cos^3x\cos 3x\right)\,dx.$$ 4. Now evaluate each term: - $$\int_0^\pi 1\,dx=\pi.$$ - $$\int_0^\pi \cos x\cos 3x\,dx=0,$$ by orthogonality. - $$\int_0^\pi \cos^2x\,dx=\frac\pi2.$$ - For $$\int_0^\pi \cos^3x\cos 3x\,dx,$$ use $$\cos 3x=4\cos^3x-3\cos x,$$ so $$\cos^3x\cos 3x=4\cos^6x-3\cos^4x.$$ Thus \begin{align*} \int_0^\pi \cos^3x\cos 3x\,dx &=4\int_0^\pi \cos^6x\,dx-3\int_0^\pi \cos^4x\,dx. \end{align*} Now the standard values are $$\int_0^\pi \cos^4x\,dx=\frac{3\pi}{8}, \qquad \int_0^\pi \cos^6x\,dx=\frac{5\pi}{16}.$$ Therefore \begin{align*} \int_0^\pi \cos^3x\cos 3x\,dx &=4\cdot \frac{5\pi}{16}-3\cdot \frac{3\pi}{8}\\ &=\frac{5\pi}{4}-\frac{9\pi}{8}\\ &=\frac{\pi}{8}. \end{align*} 5. So, \begin{align*} \int_0^\pi f(x)\,dx &=\pi+0+\frac\pi2+\frac\pi8\\ &=\frac{13\pi}{8}. \end{align*} Hence $$I=\frac12\cdot \frac{13\pi}{8}=\frac{13\pi}{16}.$$ 6. Comparing with $$I=\frac{k\pi}{16},$$ we get $$k=13.$$

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