Question

yadi bhajak 3x varg _2x+2 bhagfal​

Answer 1

Step-by-step explanation:

$\huge\mathfrak\green{\bold{\underline{☘{ ℘ɧεŋσɱεŋศɭ}☘}}}$

$\red{\bold{\underline{\underline{❥Question᎓}}}}$ Integrate the function

$\frac{ \sqrt{tanx} }{sinxcosx}$

$\huge\tt\underline\blue{「Answer」</p><p> }$

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⇛$\frac{ \sqrt{tanx} }{sinxcosx}$

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⇛$\frac{ \sqrt{tanx} }{sinxcosx \times \frac{cosx}{cosx} }$

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⇛$\frac{ \sqrt{tanx} }{sinx \times \frac{ {cos}^{2} x}{cosx} }$

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⇛$\frac{ \sqrt{tanx} }{ {cos}^{2} x \times \frac{sinx}{cosx} }$

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⇛$\frac{ \sqrt{tanx} }{ {cos}^{2}x \times tanx }$

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⇛${tan}^{ \frac{1}{2} - 1 } \times \frac{1}{ {cos}^{2} x}$

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⇛${(tan)}^{ - \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x$

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⇛${(tan)}^{ - \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = ∫ {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x \times dx$

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☛Let tanx=t

☛differentiating both sides w.r.t.x

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⇛${sec}^{2} x = \frac{dt}{dx}$

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⇛$dx \frac{dt}{ {sec}^{2}x }$

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⇛$∴∫ {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x \times dx$

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⇛$∫ {(t)}^{ - \frac{1}{2} } \times {sec}^{2} x \times \frac{dt}{ {sec}^{2}x }$

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⇛$∫ {t}^{ - \frac{1}{2} } \times dt$

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⇛$\frac{ {t}^{ - \frac{1}{2} + 1} }{ - \frac{1}{2} + 1 } + c$

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⇛$\frac{ {t}^{ \frac{1}{2} } }{ \frac{1}{2} } + c = 2 {t}^{ \frac{1}{2} } + c = 2 \sqrt{t} + c$

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⇛$2 \sqrt{t} + c = 2 \sqrt{tanx} + c$

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нσρє ıт нєłρs yσυ

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тнαηkyσυ

Answer 2

Answer:

pending tuners the exam i did not have any questions or need any gf tha aur wo wo sl vs the answer to this post