int e^{2x} frac{(sin 2x cos 2x - 1)}{sin^2 2x | vedclass

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(D) माना $I = \int e^{2x} \frac{\sin 2x \cos 2x - 1}{\sin^2 2x} \, dx$. हम समाकल्य को इस प्रकार लिख सकते हैं: $I = \int e^{2x} \left( \frac{\sin 2x \c

Vedclass · Accepted Answer

$\frac{1}{2} e^{2x} \cot(2x) + c$,जहाँ $c$ समाकलन स्थिरांक है

Vedclass · Answer

(D) माना $I = \int e^{2x} \frac{\sin 2x \cos 2x - 1}{\sin^2 2x} \, dx$. हम समाकल्य को इस प्रकार लिख सकते हैं: $I = \int e^{2x} \left( \frac{\sin 2x \cos 2x}{\sin^2 2x} - \frac{1}{\sin^2 2x} \right) \, dx$ $I = \int e^{2x} (\cot 2x - \csc^2 2x) \, dx$. अब,खंडशः समाकलन (Integration by parts) का उपयोग करते हुए: माना $u = \cot 2x$,$dv = e^{2x} \, dx$. तब $du = -2 \csc^2 2x \, dx$ और $v = \frac{1}{2} e^{2x}$. $I = \int u \, dv = uv - \int v \, du$ $I = \cot 2x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} (-2 \csc^2 2x) \, dx - \int e^{2x} \csc^2 2x \, dx$ $I = \frac{1}{2} e^{2x} \cot 2x + \int e^{2x} \csc^2 2x \, dx - \int e^{2x} \csc^2 2x \, dx + c$ $I = \frac{1}{2} e^{2x} \cot 2x + c$.

$\int e^{2x} \frac{(\sin 2x \cos 2x - 1)}{\sin^2 2x} \, dx =$

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