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∫ 1 + Cos 4 X Cot X − Tan X D X - Mathematics

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प्रश्न

\[\int\frac{1 + \cos 4x}{\cot x - \tan x} dx\]
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उत्तर

\[\int\left( \frac{1 + \cos 4x}{\cot x - \tan x} \right) dx\]

\[ = \int\frac{\left( 1 + \cos 4x \right)}{\left( \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} \right)} dx\]

\[ = \int\frac{2 \cos^2 2x \times \sin x \cos x}{\left( \cos^2 x - \sin^2 x \right)}dx\]

\[ = \int\frac{\cos^2 2x \times 2 \sin x \cos x}{\cos 2x}dx\]

\[ = \int\cos 2x  \sin 2xdx\]

\[ = \frac{1}{2}\int2 \sin 2x \cos  2xdx\]

\[ = \frac{1}{2}\int\sin 4xdx\]

\[ = \frac{1}{2}\left[ - \frac{\cos 4x}{4} \right] + C\]

\[ = - \frac{1}{8}\cos 4x + C\]

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अध्याय 19: Indefinite Integrals - Exercise 19.03 [पृष्ठ २३]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.03 | Q 16 | पृष्ठ २३

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