Math, asked by itgo16, 5 months ago

Integrate the function
 \huge\tt\frac{ \sqrt{tanx} }{sinxcosx}

Answers

Answered by Anonymous
142

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\huge{\bold☘}\mathfrak\pink{\bold{\underline{{ ℘ɧεŋσɱεŋศɭ}}}}{\bold☘}

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

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

\huge\tt{\boxed{\overbrace{\underbrace{\blue{Answer</p><p> }}}}}

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 ⇛\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} }

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⇛ \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 }

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

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

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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

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\huge\tt\bold{☛Let tanx=t}

\huge\tt\bold{☛differentiating \:both\: sides \:w.r.t.x}

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

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

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

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

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⇛ \huge\tt ∫  {t}^{ -  \frac{1}{2} }  \times dt

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

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

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

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нσρє ıт нєłρs yσυ

_____________________

тнαηkyσυ

Answered by Anonymous
0

Step-by-step explanation:

No one knows everything, but everyone knows something. With Brainly, students combine their strengths and talents to tackle problems together.

\huge{\bold☘}\mathfrak\pink{\bold{\underline{{ ℘ɧεŋσɱεŋศɭ}}}}{\bold☘}☘

℘ɧεŋσɱεŋศɭ

\red{\bold{\underline{\underline{❥Question᎓}}}}

❥Question᎓

Integrate the function

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

sinxcosx

tanx

\huge\tt{\boxed{\overbrace{\underbrace{\blue{Answer }}}}}

Answer

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

sinxcosx

tanx

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

sinxcosx×

cosx

cosx

tanx

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

sinx×

cosx

cos

2

x

tanx

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

cos

2

cosx

sinx

tanx

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

cos

2

x×tanx

tanx

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

2

1

−1

×

cos

2

x

1

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

2

1

×

cos

2

x

1

=(tanx)

2

1

×sec

2

x

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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)

2

1

×

cos

2

x

1

=∫(tanx)

2

1

×sec

2

x×dx

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\huge\tt\bold{☛Let tanx=t}☛Lettanx=t

\huge\tt\bold{☛differentiating \:both\: sides \:w.r.t.x}☛differentiatingbothsidesw.r.t.x

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

2

x=

dx

dt

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

sec

2

x

dt

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

2

1

×sec

2

x×dx

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

2

1

×sec

2

sec

2

x

dt

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⇛ \huge\tt ∫ {t}^{ - \frac{1}{2} } \times dt⇛∫t

2

1

×dt

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

2

1

+1

t

2

1

+1

+c

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

2

1

t

2

1

+c=2t

2

1

+c=2

t

+c

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

t

+c=2

tanx

+c

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нσρє ıт нєłρs yσυ

_____________________

тнαηkyσυ

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