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प्रश्न

\[\int\frac{\sin^6 x}{\cos^8 x} dx =\]

विकल्प

  •  tan 7x + C

  • \[\frac{\tan^7 x}{7} + C\]
  • \[\frac{\tan 7x}{7} + C\]
  • sec7 x + C

MCQ

उत्तर

\[\frac{\tan^7 x}{7} + C\]

\[\text{Let }I = \int\frac{\sin^6 x}{\cos^8 x}dx\]

\[ = \int\frac{\sin^6 x}{\cos^6 x} \times \frac{1}{\cos^2 x}dx\]

\[ = \int \tan^6 x \cdot \sec^2 x dx\]

\[\text{Putting }\tan x = t\]
\[ \Rightarrow \sec^2 x dx = dt\]
\[ \therefore I = \int t^6 \cdot dt\]
\[ = \frac{t^7}{7} + C\]
\[ = \frac{\tan^7 x}{7} + C ............\left( \because t = \tan x \right)\]

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अध्याय 19: Indefinite Integrals - MCQ [पृष्ठ २०१]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
MCQ | Q 15 | पृष्ठ २०१

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