Question

Please tell this answer

Answer 1

Answer:

$\orange{\bold{\underbrace{\overbrace{❥Answer᎓}}}}$

Integrate the function

$\huge\green\tt\frac{ \sqrt{tanx} }{sinxcosx}}$

⇛$\huge\tt\frac{ \sqrt{tanx} }{sinxcosx}$

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

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

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

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

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

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

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

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$\bold\blue{☛\: Let tanx=t}$

$\bold\blue{☛ \:Differentiating \: both \: sides \: w.r.t.x}$

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⇛$\huge\tt {sec}^{2} x = \frac{dt}{dx}$

⇛$\huge\tt{dx \frac{dt}{ {sec}^{2}x }}$

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

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

⇛$\huge\tt ∫ {t}^{ - \frac{1}{2} }$ㅤ ㅤ

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

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

⇛$\huge2 \sqrt{t} + c = 2 \sqrt{tanx}$

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Answer 2

Answer:

$= > \small (\frac{2}{9} )^{3} \times ( \frac{2}{9} )^{8} = ( \frac{2}{9}) ^{2x - 1} \\ \\ = > \: \: \: \: \: \: \: ( \frac{2}{9} )^{3 + 8} = ( \frac{2}{9} )^{2x - 1} \\ \\ = > \: \: \: \: \: \: \: \: \: ( \frac{2}{9} ) ^{11} = ( \frac{2}{9} ) ^{2x - 1}$

$= > 11 = 2x - 1 \\ = > 11 + 1 = 2x \\ = > \: \: \: \: \: \: \: 12 = 2x \\ = > \: \: \: \: \: \: \: \: x = \frac{12}{2} \\ = > \huge \: x = 6$

hence, the value of x = 6

Step-by-step explanation:

hope it's helpful,,

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