JEE Main 2018 — Mathematics Calculus

$$I=\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\mathrm{sin}}^{4}⁡x(1+\mathrm{ln}⁡(\frac{2+\mathrm{sin}⁡x}{2-\mathrm{sin}⁡x}))dx......(i)$$I=∫−2π​2π​​sin4⁡x(1+ln⁡(2−sin⁡x2+sin⁡x​))dx......(i)

$$I=2\int _{0}^{\frac{\pi }{2}}{\mathrm{sin}}^{4}xdx............(i),(\because {\mathrm{sin}}^{4}x\text{is an even function})$$I=2∫02π​​sin4xdx............(i),(∵sin4xis an even function)

$$(\because \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}({\mathrm{sin}}^{4}x)\mathrm{ln}⁡(\frac{2+\mathrm{sin}x}{2-\mathrm{sin}x})dx=0,\because ({\mathrm{sin}}^{4}x)\mathrm{ln}⁡(\frac{2+\mathrm{sin}x}{2-\mathrm{sin}x})\text{is odd function})$$(∵∫−2π​2π​​(sin4x)ln⁡(2−sinx2+sinx​)dx=0,∵(sin4x)ln⁡(2−sinx2+sinx​)is odd function)

Let $\displaystyle \int _{0}^{\frac{\pi }{2}}{\mathrm{sin}}^{4}x dx=m$∫02π​​sin4xdx=m

$$\Rightarrow m=\int _{0}^{\frac{\pi }{2}}{\mathrm{sin}}^{4}(\frac{\pi }{2}-x)dx=\int _{0}^{\frac{\pi }{2}}{\mathrm{cos}}^{4}(x)dx$$⇒m=∫02π​​sin4(2π​−x)dx=∫02π​​cos4(x)dx

Adding both, we get

$$\Rightarrow 2m=\int _{0}^{\frac{\pi }{2}}{\mathrm{sin}}^{4}(x)dx+\int _{0}^{\frac{\pi }{2}}{\mathrm{cos}}^{4}(x)dx$$⇒2m=∫02π​​sin4(x)dx+∫02π​​cos4(x)dx

$$\Rightarrow 2m=\int _{0}^{\frac{\pi }{2}}{{\mathrm{sin}}^{4}(x)+{\mathrm{cos}}^{4}(x)}dx$$⇒2m=∫02π​​sin4(x)+cos4(x)dx

$$\Rightarrow 2m={({\mathrm{sin}}^{2}(x)+{\mathrm{cos}}^{2}(x))}^{2}-2{\mathrm{sin}}^{2}(x){\mathrm{cos}}^{2}(x)$$⇒2m=(sin2(x)+cos2(x))2−2sin2(x)cos2(x)

$$\Rightarrow 2m=1-\frac{{\mathrm{sin}}^{2}(2x)}{2}$$⇒2m=1−2sin2(2x)​

$$\Rightarrow 2m=\int _{0}^{\frac{\pi }{2}}{1-\frac{1}{2}(\frac{1-\mathrm{cos}4x}{2})}dx=\int _{0}^{\frac{\pi }{2}}{\frac{3}{4}+\frac{\mathrm{cos}4x}{4}}dx$$⇒2m=∫02π​​1−21​(21−cos4x​)dx=∫02π​​43​+4cos4x​dx

$$\Rightarrow 2m={[\frac{3x}{4}+\frac{\mathrm{sin}4x}{16}]}_{0}^{\frac{\pi }{2}}=\frac{3\pi }{8}$$⇒2m=[43x​+16sin4x​]02π​​=83π​

$$\Rightarrow m=\frac{3\pi }{16}$$⇒m=163π​

Substituting in equation $(i)$(i), we get

$$I=2m=\frac{3\pi }{8}$$I=2m=83π​