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