Math, asked by Arshaquemirash6242, 7 months ago

Yadi varg samikarn ke mul(6+✓5) or(6-✓5) hai to varg samikarn bataiye

Answers

Answered by Anonymous
2

Answer:

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\huge\mathfrak\red{\bold{\underline{☘{ WタℜɱロղбモℜՖ}☘}}}

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

Integrate the function

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

\huge\tt{\boxed{\underbrace{\overbrace{\blue{AnSwEr:–}}}}}

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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}} 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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