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प्रश्न
\[\int\frac{1}{\sin x \cos^3 x} dx\]
बेरीज
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उत्तर
\[\int\frac{dx}{\sin x . \cos^3 x}\]
` "Dividing numerator and denominaor by " cos^4 x `
\[ = \int\frac{\frac{1}{\cos^4 x} dx}{\frac{\sin x . \cos^3 x}{\cos^4 x}}\]
` ∫ { . sec^4 x dx}/{tan x}`
` ∫ {sec^2 x . sec^2 x dx}/{tan x}`
\[ = \int\frac{\left( 1 + \tan^2 x \right) . \sec^2 x}{\tan x}dx\]
\[Let \tan x = t\]
` ⇒ sec^2 x = dx / dt`
` ⇒ sec^2 x dx = dt `
\[Now, \int\frac{\left( 1 + \tan^2 x \right) . \sec^2 x}{\tan x}dx \]
\[ = \int\frac{\left( 1 + t^2 \right)}{t}dt\]
\[ = \int\left( \frac{1}{t} + t \right)dt\]
\[ = \text{log} \left| \text{t} \right| + \frac{t^2}{2} + C\]
\[ = \text{log }\left| \tan x \right| + \frac{\tan^2 x}{2} + C\]
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