JEE Main 2024 — Mathematics Calculus

Let, $\displaystyle I={\int }_{0}^{\frac{\pi }{3}}{\mathrm{cos}}^{4}xdx$I=∫03π​​cos4xdx

$$\Rightarrow I=\frac{1}{4}{\int }_{0}^{\frac{\pi }{3}}{(2{\mathrm{cos}}^{2}x)}^{2}dx$$⇒I=41​∫03π​​(2cos2x)2dx

$$\Rightarrow I=\frac{1}{4}{\int }_{0}^{\frac{\pi }{3}}{(1+\mathrm{cos}2x)}^{2}dx$$⇒I=41​∫03π​​(1+cos2x)2dx

$$\Rightarrow I=\frac{1}{4}{\int }_{0}^{\frac{\pi }{3}}(1+{\mathrm{cos}}^{2}2x+2\mathrm{cos}2x)dx$$⇒I=41​∫03π​​(1+cos22x+2cos2x)dx

$$\Rightarrow I=\frac{1}{4}{\int }_{0}^{\frac{\pi }{3}}(1+2\mathrm{cos}2x)dx+\frac{1}{8}{\int }_{0}^{\frac{\pi }{3}}2{\mathrm{cos}}^{2}2xdx$$⇒I=41​∫03π​​(1+2cos2x)dx+81​∫03π​​2cos22xdx

$$\Rightarrow I=\frac{1}{4}{\int }_{0}^{\frac{\pi }{3}}(1+2\mathrm{cos}2x)dx+\frac{1}{8}{\int }_{0}^{\frac{\pi }{3}}(1+\mathrm{cos}4x)dx$$⇒I=41​∫03π​​(1+2cos2x)dx+81​∫03π​​(1+cos4x)dx

$$\Rightarrow I={[\frac{1}{4}(x+\mathrm{sin}2x)+\frac{1}{8}(x+\frac{\mathrm{sin}4x}{4})]}_{0}^{\frac{\pi }{3}}$$⇒I=[41​(x+sin2x)+81​(x+4sin4x​)]03π​​

$$\Rightarrow I=\frac{1}{4}(\frac{\pi }{3}+\mathrm{sin}\frac{2\pi }{3})+\frac{1}{8}(\frac{\pi }{3}+\frac{\mathrm{sin}\frac{4\pi }{3}}{4})$$⇒I=41​(3π​+sin32π​)+81​(3π​+4sin34π​​)

$$\Rightarrow I=(\frac{\pi }{12}+\frac{\sqrt{3}}{8})+(\frac{\pi }{24}-\frac{\sqrt{3}}{64})$$⇒I=(12π​+83​​)+(24π​−643​​)

$$\Rightarrow I=\frac{\pi }{8}+\frac{7\sqrt{3}}{64}$$⇒I=8π​+6473​​

So, on comparing with given value we get, $a=\frac{1}{8},b=\frac{7}{64}$a=81​,b=647​

$$\Rightarrow 9a+8b=\frac{9}{8}+\frac{7}{8}$$⇒9a+8b=89​+87​

$$\Rightarrow 9a+8b=\frac{16}{8}$$⇒9a+8b=816​

$$\Rightarrow 9a+8b=2$$⇒9a+8b=2