Question

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Answer 1

⇛\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

⇛\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)(tan)

−

2

1

×

cos

2

x

1

=(tanx)

−

2

1

×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)(tan)

−

2

1

×

cos

2

x

1

=∫(tanx)

−

2

1

×sec

2

x×dx⇛(tan)

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\bold\blue{☛\: Let tanx=t}☛Lettanx=t

\bold\blue{☛ \: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

⇛\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

⇛\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

⇛\huge\tt ∫ {t}^{ - \frac{1}{2} }∫t

−

2

1

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

−

2

1

+1

t

−

2

1

+1

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

2

1

t

2

1

+c=2t

2

1

+c=2

t

⇛\huge2 \sqrt{t} + c = 2 \sqrt{tanx}2

t

+c=2

tanx

Answer 2

Answer:darmowe      0

 50

Step-by-step explanation: