### Related Formula int frac1u^2 + a^2 du = frac1a tan^-1left(fracuaright) + C$\int \frac{1}{u^2 + a^2} du = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$ ### Core Logic Simplify the given integrand by converting all trigonometric ratios into sin x$\sin x$ and cos x$\cos x$: I = fracoperatornamecosec x + sin xoperatornamecosec x sec x + tan x sin^2 x$I = \frac{\operatorname{cosec} x + \sin x}{\operatorname{cosec} x \sec x + \tan x \sin^2 x}$ Numerator: frac1sin x + sin x = frac1 + sin^2 xsin x$\frac{1}{\sin x} + \sin x = \frac{1 + \sin^2 x}{\sin x}$ Denominator: frac1sin x cos x + fracsin xcos x cdot sin^2 x = frac1 + sin^4 xsin x cos x$\frac{1}{\sin x \cos x} + \frac{\sin x}{\cos x} \cdot \sin^2 x = \frac{1 + \sin^4 x}{\sin x \cos x}$ Divide Numerator by Denominator: I = fracfrac1 + sin^2 xsin xfrac1 + sin^4 xsin x cos x = frac(1 + sin^2 x)cos x1 + sin^4 x$I = \frac{\frac{1 + \sin^2 x}{\sin x}}{\frac{1 + \sin^4 x}{\sin x \cos x}} = \frac{(1 + \sin^2 x)\cos x}{1 + \sin^4 x}$ ### Step 1: Apply Substitution Now rewrite the integral: y(x) = int frac(1 + sin^2 x)cos x1 + sin^4 x dx$y(x) = \int \frac{(1 + \sin^2 x)\cos x}{1 + \sin^4 x} dx$ Let sin x = t$\sin x = t$, then cos x \, dx = dt$\cos x \, dx = dt$. The integral becomes: y(x) = int frac1 + t^21 + t^4 dt$y(x) = \int \frac{1 + t^2}{1 + t^4} dt$ Divide numerator and denominator by t^2$t^2$: y(x) = int frac1 + frac1t^2t^2 + frac1t^2 dt$y(x) = \int \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} dt$ We know that t^2 + frac1t^2 = left(t - frac1tright)^2 + 2$t^2 + \frac{1}{t^2} = \left(t - \frac{1}{t}\right)^2 + 2$. Let u = t - frac1t$u = t - \frac{1}{t}$, then du = left(1 + frac1t^2right) dt$du = \left(1 + \frac{1}{t^2}\right) dt$. y(x) = int fracduu^2 + (sqrt2)^2 = frac1sqrt2tan^-1left(fracusqrt2right) + C$y(x) = \int \frac{du}{u^2 + (\sqrt{2})^2} = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{u}{\sqrt{2}}\right) + C$ y(x) = frac1sqrt2tan^-1left(fract - 1/tsqrt2right) + C = frac1sqrt2tan^-1left(fracsin x - operatornamecosec xsqrt2right) + C$y(x) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t - 1/t}{\sqrt{2}}\right) + C = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sin x - \operatorname{cosec} x}{\sqrt{2}}\right) + C$ ### Step 2: Solve for Constant C We are given that lim_xrightarrow(pi/2)^- y(x) = 0$\lim_{x\rightarrow(\pi/2)^-} y(x) = 0$. As x to fracpi2$x \to \frac{\pi}{2}$, sin x to 1$\sin x \to 1$ and operatornamecosec x to 1$\operatorname{cosec} x \to 1$. Thus, sin x - operatornamecosec x to 0$\sin x - \operatorname{cosec} x \to 0$. 0 = frac1sqrt2tan^-1left(frac1 - 1sqrt2right) + C$0 = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{1 - 1}{\sqrt{2}}\right) + C$ 0 = frac1sqrt2tan^-1(0) + C Rightarrow C = 0$0 = \frac{1}{\sqrt{2}}\tan^{-1}(0) + C \Rightarrow C = 0$ ### Step 3: Find y(pi/4) Substitute x = fracpi4$x = \frac{\pi}{4}$ into the exact solution y(x) = frac1sqrt2tan^-1left(fracsin x - operatornamecosec xsqrt2right)$y(x) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sin x - \operatorname{cosec} x}{\sqrt{2}}\right)$: At x = fracpi4$x = \frac{\pi}{4}$, sinleft(fracpi4right) = frac1sqrt2$\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and operatornamecosecleft(fracpi4right) = sqrt2$\operatorname{cosec}\left(\frac{\pi}{4}\right) = \sqrt{2}$. yleft(fracpi4right) = frac1sqrt2tan^-1left(fracfrac1sqrt2 - sqrt2sqrt2right)$y\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\frac{1}{\sqrt{2}} - \sqrt{2}}{\sqrt{2}}\right)$ = frac1sqrt2tan^-1left(fracfrac1 - 2sqrt2sqrt2right) = frac1sqrt2tan^-1left(frac-1/sqrt2sqrt2right)$= \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\frac{1 - 2}{\sqrt{2}}}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{-1/\sqrt{2}}{\sqrt{2}}\right)$ = frac1sqrt2tan^-1left(-frac12right)$= \frac{1}{\sqrt{2}}\tan^{-1}\left(-\frac{1}{2}\right)$ ### Pattern Recognition A high-degree polynomial of sin x$\sin x$ or cos x$\cos x$ in the denominator matched with their derivative elements on top is the hallmark of the t+1/t$t+1/t$ or t-1/t$t-1/t$ structural substitution form. Dividing out the middle power (t^2$t^2$) aligns the differential instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus