If ∫ _ 0 ^ π 5 ^ x (1 + x 3x + ^ 2 x + ^ 3 xcos3x)dx 1 + 5 ^ x =...

Given,

$I = {∫}_{0}^{\pi}\dfrac{{5}^{\cos x}\left(1 + \cos x \cos 3x + \cos^{2}x + \cos^{3}xcos3x\right)}{1 + {5}^{\cos x}}dx . . . . \left(1\right)$

Now using the property ${∫}_{a}^{b}f\left(x\right) d x = {∫}_{a}^{b}f\left(a + b - x\right) d x$ we get,

$⇒ I = {∫}_{0}^{\pi}\dfrac{{5}^{- \cos x}\left(1 + \cos x \cos 3x + \cos^{2}x + \cos^{3}xcos3x\right)}{1 + {5}^{- \cos x}}dx . . . . \left(2\right)$

Now adding above equations we get,

$2I = {∫}_{0}^{\pi}\left(1 + \cos x \cos 3x + \cos^{2}x + \cos^{3}xcos3x\right)dx$

Now using the property ${∫}_{0}^{2a}f\left(x\right) d x = 2{∫}_{a}^{a}f\left(x\right) d x\left\{\text{when}f\left(x\right)\text{is even}\right\}$

$⇒ 2I = 2{∫}_{0}^{\dfrac{\pi}{2}}\left(1 + \cos x \cos 3x + \cos^{2}x + \cos^{3}xcos3x\right)dx$

$⇒ I = {∫}_{0}^{\dfrac{\pi}{2}}\left(1 + \cos x \cos 3x + \cos^{2}x + \cos^{3}xcos3x\right)dx . . . . . \left(3\right)$

$⇒ I = {∫}_{0}^{\dfrac{\pi}{2}}\left(1 + \sin x\left(- \sin 3x\right) + \sin^{2}x - \sin^{3}xsin3x\right)dx . . . . \left(4\right)$

Now adding equation \left(3\right) \text{ and } \left(4\right)$we get,</p><p>$⇒ 2I = {∫}_{0}^{\dfrac{\pi}{2}}\left(3 + \cos 4x + \cos^{3}xcos3x - \sin^{3}xsin3x\right)dx$</p><p>$⇒ 2I = {∫}_{0}^{\dfrac{\pi}{2}}3 + \cos 4x + \left(\dfrac{\cos 3x + 3\cos x}{4}\right)\cos 3x - \sin 3x\left(\dfrac{3\sin x - \sin 3x}{4}\right)dx$</p><p>$⇒ 2I = {∫}_{0}^{\dfrac{\pi}{2}}\left(3 + \cos 4x + \dfrac{1}{4} + \dfrac{3}{4}\cos 4x\right)dx$</p><p>$⇒ 2I = \dfrac{13}{4} \times \dfrac{\pi}{2} + \dfrac{7}{4}{\left(\dfrac{\sin 4x}{4}\right)}_{0}^{\dfrac{\pi}{2}}$</p><p>$⇒ I = \dfrac{13\pi}{16}$,</p><p>Now on comparing with$\dfrac{k\pi}{16}$, we get$k = 13$