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∫ √ Tan X Sin X Cos X D X - Mathematics

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प्रश्न

\[\int\frac{\sqrt{\tan x}}{\sin x \cos x} dx\]

योग
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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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अध्याय 19: Indefinite Integrals - Exercise 19.09 [पृष्ठ ५८]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.09 | Q 16 | पृष्ठ ५८

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