Question

tan-' (1)+cot-' (1)+sin*'(-1).

Answer 1

Step-by-step explanation:

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

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

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

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

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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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☛Let tanx=t

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

╚════════════════════════╝

нσρє ıт нєłρs yσυ

_____________________

тнαηkyσυ

Answer 2

Answer:

Step-by-step explanation:

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

☘℘ɧεŋσɱεŋศɭ☘

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

❥Question᎓

Integrate the function

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

sinxcosx

tanx

\huge\tt\underline\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

x×

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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☛Let tanx=t

☛differentiating both sides w.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

x×

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

╚════════════════════════╝

нσρє ıт нєłρs yσυ

_____________________

тнαηkyσυ