Definite Integration Question 24
Question 24 - 01 February - Shift 2
If $\int_0^{\pi} \frac{5^{\cos x}(1+\cos x \cos 3 x+\cos ^{2} x+\cos ^{3} x \cos 3 x) d x}{1+5^{\cos x}}=\frac{k \pi}{16}$, then $k$ is equal to
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Answer: 13
Solution:
Formula: Properties of definite integral, Triple Angle Identities
$ \begin{aligned} & I=\int_0^{\pi} \frac{5^{\cos x}(1+\cos x \cos 3 x+\cos ^{2} x+\cos ^{3} x \cos 3 x)}{1+5^{\cos x}} d x \ldots (1)\\ & I=\int_0^{\pi} \frac{5^{-\cos x}(1+\cos x \cos 3 x+\cos ^{2} x+\cos ^{3} x \cos 3 x)}{1+5^{-\cos x}} d x \ldots (2)\\ & on \ adding \ (1) \ and \ (2) \\ & 2 I=\int_0^{\pi}(1+\cos x \cos 3 x+\cos ^{2} x+\cos ^{3} x \cos 3 x) d x \\ & \not 2 I=\not 2 \int_0^{\frac{\pi}{2}}(1+\cos x \cos 3 x+\cos ^{2} x+\cos ^{3} x \cos 3 x) d x \\ & I= \int_0^{\frac{\pi}{2}}(1+\cos x \cos 3 x+\cos ^{2} x+\cos ^{3} x \cos 3 x) d x \ldots (3) \\ & I=\int_0^{\frac{\pi}{2}}(1+\sin x(-\sin 3 x)+\sin ^{2} x-\sin ^{3} x \sin 3 x) d x \ldots (4)\\ & 2 I=\int_0^{\frac{\pi}{2}}(3+\cos 4 x+\cos ^{3} x \cos 3 x-\sin ^{3} x \sin 3 x) d x \\ & 2 I=\int_0^{\frac{\pi}{2}} 3+\cos 4 x+(\frac{\cos 3 x+3 \cos x}{4}) \cos 3 x-\sin 3 x(\frac{3 \sin x-\sin 3 x}{4}) d x \\ & 2 I=\int_0^{\frac{\pi}{2}}(3+\cos 4 x+\frac{1}{4}+\frac{3}{4} \cos 4 x) d x \\ & 2 I=\frac{13}{4} \times \frac{\pi}{2}+\frac{7}{4}(\frac{\sin 4 x}{4})_0^{\frac{\pi}{2}} \\ & \Rightarrow I=\frac{13 \pi}{16}=\frac{k \pi}{16} \end{aligned} $
On comparing, we get $k=13$
