If 0^ 5^ x (1 + x 3x + ^2 x + ^3 x 3x)1 + 5^ x dx = k 16, then k is equal to.

Given: I= 0^ 5^ x(1+ x 3x+ ^2x+ ^3x 3x)1+5^ x dx Find: k if I= k 16. Using the symmetry substitution x -x, I= 0^ 5^- x(1+ x 3x+ ^2x+ ^3x 3x)1+5^- x dx Adding the two forms, 2I= 0^ (1+ x 3x+ ^2x+ ^3x 3x) dx Now use symmetry about 2: 2I=2 0^ 2(1+ x 3x+ ^2x+ ^3x 3x) dx Equivalently, I= 0^ 2(1- x 3x+ ^2x- ^3x 3x) dx Adding the two half-interval expressions, 2I= 0^ 2(3+ 4x+ ^3x 3x- ^3x 3x) dx Using ^3x= 3x+3 x4, ^3x= 3 x- 3x4 we get 2I= 0^ 2 (3+ 4x+ 3x( 3x+3 x)4- 3x(3 x- 3x)4)dx so 2I= 0^ 2 (3+ 4x+ 14+ 34 4x)dx Hence, 2I= 134 2+ 74 ( 4x4)0^ 2 Since the sine term is 0, 2I= 13 8 and therefore I= 13 16 Comparing with I= k 16, we get k=13. Therefore, the value of k is 13. The solution also displays 26 in one place, but the worked steps conclude I= 13 16, so the correct value is 13.

If 0^ 5^ x (1 + x 3x + ^2 x + ^3 x 3x)1 + 5^ x dx = k 16, then k… — Solution | JEE Mathematics

NVAMediumJEE 2023Properties of Definite Integrals

If $\int_0^\pi \frac{5^{\cos x} (1 + \cos x \cos 3x + \cos^2 x + \cos^3 x \cos 3x)}{1 + 5^{\cos x}} \, dx = \frac{k \pi}{16}$ , then $k$ is equal to _____.

Answer

Correct answer:13

Step-by-step solution

Standard Method

Given:

I=\int_{0}^{\pi}\frac{5^{\cos x}(1+\cos x\cos 3x+\cos^{2}x+\cos^{3}x\cos 3x)}{1+5^{\cos x}}\,dx

Find: $k$ if $I=\frac{k\pi}{16}$ .

Using the symmetry substitution $x\to \pi-x$ ,

I=\int_{0}^{\pi}\frac{5^{-\cos x}(1+\cos x\cos 3x+\cos^{2}x+\cos^{3}x\cos 3x)}{1+5^{-\cos x}}\,dx

Adding the two forms,

2I=\int_{0}^{\pi}(1+\cos x\cos 3x+\cos^{2}x+\cos^{3}x\cos 3x)\,dx

Now use symmetry about $\frac{\pi}{2}$ :

2I=2\int_{0}^{\frac{\pi}{2}}(1+\cos x\cos 3x+\cos^{2}x+\cos^{3}x\cos 3x)\,dx

Equivalently,

I=\int_{0}^{\frac{\pi}{2}}(1-\sin x\sin 3x+\sin^{2}x-\sin^{3}x\sin 3x)\,dx

Adding the two half-interval expressions,

2I=\int_{0}^{\frac{\pi}{2}}(3+\cos 4x+\cos^{3}x\cos 3x-\sin^{3}x\sin 3x)\,dx

Using

\cos^{3}x=\frac{\cos 3x+3\cos x}{4}, \qquad \sin^{3}x=\frac{3\sin x-\sin 3x}{4}

we get

2I=\int_{0}^{\frac{\pi}{2}}\left(3+\cos 4x+\frac{\cos 3x(\cos 3x+3\cos x)}{4}-\frac{\sin 3x(3\sin x-\sin 3x)}{4}\right)dx

2I=\int_{0}^{\frac{\pi}{2}}\left(3+\cos 4x+\frac{1}{4}+\frac{3}{4}\cos 4x\right)dx

Hence,

2I=\frac{13}{4}\cdot\frac{\pi}{2}+\frac{7}{4}\left(\frac{\sin 4x}{4}\right)_{0}^{\frac{\pi}{2}}

Since the sine term is $0$ ,

2I=\frac{13\pi}{8}

and therefore

I=\frac{13\pi}{16}

Comparing with $I=\frac{k\pi}{16}$ , we get $k=13$ .

Therefore, the value of $k$ is $13$ .

The solution also displays 26 in one place, but the worked steps conclude $I=\frac{13\pi}{16}$ , so the correct value is $13$ .

Common mistakes

Using the displayed 26 from the solution without checking the working is incorrect. The derivation ends with $I=\frac{13\pi}{16}$ , so comparing with $\frac{k\pi}{16}$ gives $k=13$ , not $26$ .
While applying the substitution $x\to \pi-x$ , forgetting that $\cos(\pi-x)=-\cos x$ and $\cos 3(\pi-x)=-\cos 3x$ leads to wrong signs. Recompute each transformed trigonometric term carefully before adding the two integrals.
After obtaining $2I$ , dividing incorrectly at the final step is a common error. If $2I=\frac{13\pi}{8}$ , then $I=\frac{13\pi}{16}$ . Always halve the entire expression before comparing coefficients.

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