Math, asked by Anonymous, 5 months ago

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

Answers

Answered by Anonymous
19

Step-by-step explanation:

Step-by-step explanation:

\orange{\bold{\underbrace{\overbrace{❥Question᎓}}}}

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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Answered by BʀᴀɪɴʟʏAʙCᴅ
4

\huge\mathcal{\mid{\mid{\underline{\pink{Good\:Morning\:}}}{\mid{\mid}}}} \\

\Large{\colorbox{pink}{\bf{\green{QuEsTiOn;-}}}} \\

\red\checkmark\:\:\bf\blue{Integral\:of\:\dfrac{\sqrt{tanx}}{sinx.cosx}} \\

\huge{\orange{\boxed{\fcolorbox{lime}{indigo}{\color{aqua}ANSWER}}}} \\

:\implies\:\:\Large\bf{\int\:\dfrac{\sqrt{tanx}}{sinx.cosx}\:dx} \\

:\implies\:\:\bf{\int\:\dfrac{\sqrt{tanx}}{(\frac{sinx.cosx}{cos^2x})\:cos^2x}\:dx} \\

:\implies\:\:\bf{\int\:\dfrac{\sqrt{tanx}}{(\frac{sinx}{cosx})\:cos^2x}\:dx} \\

:\implies\:\:\bf{\int\:\dfrac{\sqrt{tanx}}{tanx}\:.\:sec^2x\:dx} \\

:\implies\:\:\bf{\int\:\dfrac{1}{\sqrt{tanx}}\:.\:sec^2x\:dx} \\

\Large\bf\pink{Let} \\

  • \bf{tanx\:=\:t}

\longmapsto\:\:\bf{sec^2x\:.\:dx\:=\:dt\:} \\

:\implies\:\:\bf{\int\:\dfrac{1}{t^{1/2}}\:.\:dt} \\

:\implies\:\:\bf{\int\:t^{-\:\frac{1}{2}}\:.\:dt} \\

:\implies\:\:\bf{\dfrac{t^{-\:\frac{1}{2}\:+\:1}}{-\:\frac{1}{2}\:+\:1}\:+\:c} \\

:\implies\:\:\bf{\dfrac{t^{\frac{1}{2}}}{\frac{1}{2}}\:+\:c} \\

:\implies\:\:\bf{2\:t^{\frac{1}{2}}\:+\:c} \\

:\implies\:\:\bf\green{2\:\sqrt{tanx}\:+\:c} \\

\Large\mathbb\pink{THANKS} \\

\Large\mathbb\green{HOPE\: IT'S\: HELPFUL} \\

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