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प्रश्न
\[\int\frac{\sqrt{\tan x}}{\sin x \cos x} dx\]
बेरीज
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उत्तर
\[\int\frac{\sqrt{\tan x}}{\sin x \cos x}dx\]
\[ = \int\frac{\sqrt{\tan x}}{\frac{\sin x}{\cos x} \times \cos^2 x}dx\]
\[ = \int\frac{\sqrt{\tan x}}{\tan x} \times \sec^2 \text{x dx}\]
\[ = \int\frac{1}{\sqrt{\tan x}} \times \sec^2 \text{x dx}\]
\[ = \int \left( \tan x \right)^{- \frac{1}{2}} \sec^2 x dx\]
\[Let \tan x = t\]
\[ \Rightarrow \sec^2 x = \frac{dt}{dx}\]
\[ \Rightarrow \sec^2 \text{x dx} = dt\]
\[Now, \int \left( \tan x \right)^{- \frac{1}{2}} \sec^2\text{ x dx}\]
\[ = \int t^{- \frac{1}{2}} dt\]
\[ = \left[ \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + C\]
\[ =\text{2} \sqrt{t} + C\]
\[ = 2 \sqrt{\tan x} + C\]
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