If I = int frac{sin^2 x - 1}{2x sin^2 x + sin | vedclass

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(A) Given $I = \int \frac{\sin^2 x - 1}{2x \sin^2 x + \sin 2x} dx$. Since $\sin^2 x - 1 = -\cos^2 x$ and $\sin 2x = 2 \sin x \cos x$,we have: $I = \in

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$ce^{2I} \sin x = x \sin x + \cos x$

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(A) Given $I = \int \frac{\sin^2 x - 1}{2x \sin^2 x + \sin 2x} dx$. Since $\sin^2 x - 1 = -\cos^2 x$ and $\sin 2x = 2 \sin x \cos x$,we have: $I = \int \frac{-\cos^2 x}{2x \sin^2 x + 2 \sin x \cos x} dx = \int \frac{-\cos^2 x}{2 \sin x (x \sin x + \cos x)} dx$. Divide numerator and denominator by $\sin^2 x$: $I = \int \frac{-\cot^2 x}{2(x + \cot x)} dx$. Let $u = x + \cot x$,then $du = (1 - \csc^2 x) dx = (1 - (1 + \cot^2 x)) dx = -\cot^2 x dx$. Substituting these into the integral: $I = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|u| + c = \frac{1}{2} \ln|x + \cot x| + c$. Thus,$2I = \ln|x + \cot x| + 2c$. Taking the exponential of both sides: $e^{2I} = e^{\ln|x + \cot x| + 2c} = |x + \cot x| e^{2c}$. Let $C = e^{2c}$,then $Ce^{2I} = x + \cot x = x + \frac{\cos x}{\sin x} = \frac{x \sin x + \cos x}{\sin x}$. Therefore,$Ce^{2I} \sin x = x \sin x + \cos x$.

If $I = \int \frac{\sin^2 x - 1}{2x \sin^2 x + \sin 2x} dx$,then. . . . . $( \sin x \neq 0)$ (where $c$ is a constant of integration)

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