JEE Main 2024 — Mathematics Calculus

Given:

$$f(x)=\int \frac{\mathrm{cosec}x+\mathrm{sin}x}{\mathrm{cosec}x\mathrm{sec}x+\mathrm{tan}x{\mathrm{sin}}^{2}x}dx$$f(x)=∫cosecxsecx+tanxsin2xcosecx+sinx​dx

$$\Rightarrow f(x)=\int \frac{\frac{1}{\mathrm{sin}x}+\mathrm{sin}x}{\frac{1}{\mathrm{sin}x\mathrm{cos}x}+\frac{\mathrm{sin}x}{\mathrm{cos}x}({\mathrm{sin}}^{2}x)}dx$$⇒f(x)=∫sinxcosx1​+cosxsinx​(sin2x)sinx1​+sinx​dx

$$\Rightarrow f(x)=\int \frac{\frac{1+{\mathrm{sin}}^{2}x}{\mathrm{sin}x}}{\frac{1+{\mathrm{sin}}^{4}x}{\mathrm{sin}x\mathrm{cos}x}}dx$$⇒f(x)=∫sinxcosx1+sin4x​sinx1+sin2x​​dx

$$\Rightarrow f(x)=\int \frac{(1+{\mathrm{sin}}^{2}x)\mathrm{cos}x}{1+{\mathrm{sin}}^{4}x}dx$$⇒f(x)=∫1+sin4x(1+sin2x)cosx​dx

Let, $\mathrm{sin}x=t$sinx=t

$$\mathrm{cos}xdx=dt$$cosxdx=dt

$$\Rightarrow f(x)=\int \frac{1+{t}^{2}}{1+{t}^{4}}dt$$⇒f(x)=∫1+t41+t2​dt

$$\Rightarrow f(x)=\int \frac{1+\frac{1}{{t}^{2}}}{{t}^{2}+\frac{1}{{t}^{2}}}dt$$⇒f(x)=∫t2+t21​1+t21​​dt

$$\Rightarrow f(x)=\int \frac{1+\frac{1}{{t}^{2}}}{{(t-\frac{1}{t})}^{2}+2}dt$$⇒f(x)=∫(t−t1​)2+21+t21​​dt

Let, $t-\frac{1}{t}=u$t−t1​=u

$$\Rightarrow f(x)=\int \frac{1}{{u}^{2}+2}du$$⇒f(x)=∫u2+21​du

$$\Rightarrow f(x)=\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}(\frac{u}{\sqrt{2}})+C$$⇒f(x)=2​1​tan−1(2​u​)+C

$$\Rightarrow f(x)=\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}(\frac{t-\frac{1}{t}}{\sqrt{2}})+C$$⇒f(x)=2​1​tan−1(2​t−t1​​)+C

$$\Rightarrow f(x)=\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}(\frac{\mathrm{sin}x-\frac{1}{\mathrm{sin}x}}{\sqrt{2}})+C$$⇒f(x)=2​1​tan−1(2​sinx−sinx1​​)+C

$$\Rightarrow f(x)=\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}(\frac{\mathrm{sin}x-\mathrm{cosec}x}{\sqrt{2}})+C$$⇒f(x)=2​1​tan−1(2​sinx−cosecx​)+C

Now, $\underset{x\rightarrow {\frac{\pi }{2}}^{-}}{\mathrm{lim}}f(x)=0$x→2π​−lim​f(x)=0

$$\Rightarrow \underset{x\rightarrow {\frac{\pi }{2}}^{-}}{\mathrm{lim}}f(x)=\underset{x\rightarrow {\frac{\pi }{2}}^{-}}{\mathrm{lim}}[\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}(\frac{\mathrm{sin}x-\mathrm{cosec}x}{\sqrt{2}})+C]=0$$⇒x→2π​−lim​f(x)=x→2π​−lim​[2​1​tan−1(2​sinx−cosecx​)+C]=0

$$\Rightarrow \frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}(0)+C=0$$⇒2​1​tan−1(0)+C=0

$$\Rightarrow C=0$$⇒C=0

$$\Rightarrow f(x)=\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}(\frac{\mathrm{sin}x-\mathrm{cosec}x}{\sqrt{2}})$$⇒f(x)=2​1​tan−1(2​sinx−cosecx​)

$$\Rightarrow y(\frac{\pi }{4})=\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}(\frac{\mathrm{sin}\frac{\pi }{4}-\mathrm{cosec}\frac{\pi }{4}}{\sqrt{2}})$$⇒y(4π​)=2​1​tan−1(2​sin4π​−cosec4π​​)

$$\Rightarrow y(\frac{\pi }{4})=\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}(\frac{\frac{1}{\sqrt{2}}-\sqrt{2}}{\sqrt{2}})$$⇒y(4π​)=2​1​tan−1(2​2​1​−2​​)

$$\Rightarrow y(\frac{\pi }{4})=\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}(\frac{\frac{-1}{\sqrt{2}}}{\sqrt{2}})$$⇒y(4π​)=2​1​tan−1(2​2​−1​​)

$$\Rightarrow y(\frac{\pi }{4})=\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}(\frac{-1}{2})$$⇒y(4π​)=2​1​tan−1(2−1​)