JEE Main 2023 — Mathematics Calculus

Given,

$$I={\int }_{0}^{\pi }\frac{{5}^{\mathrm{cos}x}(1+\mathrm{cos}x\mathrm{cos}3x+{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{3}x\mathrm{cos}3x)}{1+{5}^{\mathrm{cos}x}}dx....(1)$$I=∫0π​1+5cosx5cosx(1+cosxcos3x+cos2x+cos3xcos3x)​dx....(1)

Now using the property $\displaystyle {\int }_{a}^{b}f(x)dx={\int }_{a}^{b}f(a+b-x)dx$∫ab​f(x)dx=∫ab​f(a+b−x)dx we get,

$$\Rightarrow I={\int }_{0}^{\pi }\frac{{5}^{-\mathrm{cos}x}(1+\mathrm{cos}x\mathrm{cos}3x+{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{3}x\mathrm{cos}3x)}{1+{5}^{-\mathrm{cos}x}}dx....(2)$$⇒I=∫0π​1+5−cosx5−cosx(1+cosxcos3x+cos2x+cos3xcos3x)​dx....(2)

Now adding above equations we get,

$$2I={\int }_{0}^{\pi }(1+\mathrm{cos}x\mathrm{cos}3x+{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{3}x\mathrm{cos}3x)dx$$2I=∫0π​(1+cosxcos3x+cos2x+cos3xcos3x)dx

Now using the property $\displaystyle {\int }_{0}^{2a}f(x)dx=2{\int }_{a}^{a}f(x)dx{\text{when }f(x)\text{ is even}}$∫02a​f(x)dx=2∫aa​f(x)dxwhen f(x) is even

$$\Rightarrow 2I=2{\int }_{0}^{\frac{\pi }{2}}(1+\mathrm{cos}x\mathrm{cos}3x+{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{3}x\mathrm{cos}3x)dx$$⇒2I=2∫02π​​(1+cosxcos3x+cos2x+cos3xcos3x)dx

$$\Rightarrow I={\int }_{0}^{\frac{\pi }{2}}(1+\mathrm{cos}x\mathrm{cos}3x+{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{3}x\mathrm{cos}3x)dx.....(3)$$⇒I=∫02π​​(1+cosxcos3x+cos2x+cos3xcos3x)dx.....(3)

$$\Rightarrow I={\int }_{0}^{\frac{\pi }{2}}(1+\mathrm{sin}x(-\mathrm{sin}3x)+{\mathrm{sin}}^{2}x-{\mathrm{sin}}^{3}x\mathrm{sin}3x)dx....(4)$$⇒I=∫02π​​(1+sinx(−sin3x)+sin2x−sin3xsin3x)dx....(4)

Now adding equation (3)&(4) we get,

$$\Rightarrow 2I={\int }_{0}^{\frac{\pi }{2}}(3+\mathrm{cos}4x+{\mathrm{cos}}^{3}x\mathrm{cos}3x-{\mathrm{sin}}^{3}x\mathrm{sin}3x)dx$$⇒2I=∫02π​​(3+cos4x+cos3xcos3x−sin3xsin3x)dx

$$\Rightarrow 2I={\int }_{0}^{\frac{\pi }{2}}3+\mathrm{cos}4x+(\frac{\mathrm{cos}3x+3\mathrm{cos}x}{4})\mathrm{cos}3x-\mathrm{sin}3x(\frac{3\mathrm{sin}x-\mathrm{sin}3x}{4})dx$$⇒2I=∫02π​​3+cos4x+(4cos3x+3cosx​)cos3x−sin3x(43sinx−sin3x​)dx

$$\Rightarrow 2I={\int }_{0}^{\frac{\pi }{2}}(3+\mathrm{cos}4x+\frac{1}{4}+\frac{3}{4}\mathrm{cos}4x)\mathrm{dx}$$⇒2I=∫02π​​(3+cos4x+41​+43​cos4x)dx

$$\Rightarrow 2I=\frac{13}{4}\times \frac{\pi }{2}+\frac{7}{4}{(\frac{\mathrm{sin}4x}{4})}_{0}^{\frac{\pi }{2}}$$⇒2I=413​×2π​+47​(4sin4x​)02π​​

$\Rightarrow I=\frac{13\pi }{16}$⇒I=1613π​,

Now on comparing with $\frac{k\pi }{16}$16kπ​, we get $k=13$k=13