मराठी

Evaluate: ∫ Cos 2 X + 2 Sin 2 X Sin 2 X D X

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प्रश्न

Evaluate:

\[\int\frac{\cos 2x + 2 \sin^2 x}{\sin^2 x}dx\]
बेरीज
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उत्तर

\[\int\left( \frac{\cos 2x + 2 \sin^2 x}{\sin^2 x} \right)dx\]
\[ = \int\left( \frac{1 - 2 \sin^2 x + 2 \sin^2 x}{\sin^2 x} \right)dx \left[ \because \cos 2x = 1 - 2 \sin^2 x \right]\]
\[ = \int {cosec}^2\text{ x   dx}\]
\[ = - \cot x + C\]

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पाठ 18: Indefinite Integrals - Exercise 19.01 [पृष्ठ ४]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 18 Indefinite Integrals
Exercise 19.01 | Q 5.1 | पृष्ठ ४

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