7.As x approaches a, evaluate: [tex] {x}^{n } + a {x}^{n – 1} + {a}^{2} {x}^{n – 2} + … + {a}^{n} [/tex] About the author Genesis
Answer: tex]\orange{\bold{\underbrace{\overbrace{❥Answer᎓}}}}[/tex] Integrate the function [tex]\huge\green\tt\frac{ \sqrt{tanx} }{sinxcosx}}[/tex] ⇛[tex]\huge\tt\frac{ \sqrt{tanx} }{sinxcosx}[/tex] ㅤ ㅤ ㅤ ㅤ ㅤ ⇛[tex]\huge\tt \frac{ \sqrt{tanx} }{sinxcosx \times \frac{cosx}{cosx}}[/tex] ㅤ ㅤ ㅤ ㅤ ㅤ ⇛[tex]\huge\tt \frac{ \sqrt{tanx} }{sinx \times \frac{ {cos}^{2} x}{cosx}}[/tex] ㅤ ㅤ ㅤ ⇛ [tex]\huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2} x \times \frac{sinx}{cosx} }[/tex] ㅤ ㅤ ㅤ ㅤ ㅤ ⇛[tex]\huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2}x \times tanx }[/tex] ⇛[tex]\huge\tt {tan}^{ \frac{1}{2} – 1 } \times \frac{1}{ {cos}^{2} x}[/tex]ㅤ ㅤ ㅤ ㅤ ㅤ ⇛[tex]\huge\tt {(tan)}^{ – \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = {(tanx)}^{ – \frac{1}{2} } \times {sec}^{2} x⇛(tan)[/tex] ㅤ ㅤ ㅤ ㅤ ㅤ ⇛[tex]\huge\tt {(tan)}^{ – \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = ∫ {(tanx)}^{ – \frac{1}{2} } \times {sec}^{2} x \times dx⇛(tan)[/tex] ㅤ ㅤ ㅤ ㅤ ㅤ [tex]\bold\blue{☛\: Let tanx=t}[/tex] [tex]\bold\blue{☛ \:Differentiating \: both \: sides \: w.r.t.x}[/tex] ㅤ ㅤ ㅤ ㅤ ㅤ ⇛[tex]\huge\tt {sec}^{2} x = \frac{dt}{dx}[/tex] ⇛[tex]\huge\tt{dx \frac{dt}{ {sec}^{2}x }}[/tex] ㅤ ㅤ ㅤ ㅤ ㅤ ⇛[tex]\huge\tt∴∫ {(tanx)}^{ – \frac{1}{2} } \times {sec}^{2} x \times dx[/tex] ⇛[tex]\huge\tt ∫ {(t)}^{ – \frac{1}{2} } \times {sec}^{2} x \times \frac{dt}{ {sec}^{2}x }[/tex] ⇛[tex]\huge\tt ∫ {t}^{ – \frac{1}{2} }[/tex]ㅤ ㅤ ⇛ [tex]\huge\tt\frac{ {t}^{ – \frac{1}{2} + 1} }{ – \frac{1}{2} + 1 }[/tex] ⇛ [tex]\huge\tt \frac{ {t}^{ \frac{1}{2} } }{ \frac{1}{2} } + c = 2 {t}^{ \frac{1}{2} } + c = 2 \sqrt{t}[/tex] ⇛[tex]\huge2 \sqrt{t} + c = 2 \sqrt{tanx}[/tex] ╚════════════ Reply
Answer: [tex]⇛\huge2 \sqrt{t} + c = 2 \sqrt{tanx}2t+c=2tanx ╚════════════ [/tex] is the answer. hope it helps you. thanks. Reply
Answer:
tex]\orange{\bold{\underbrace{\overbrace{❥Answer᎓}}}}[/tex]
Integrate the function
[tex]\huge\green\tt\frac{ \sqrt{tanx} }{sinxcosx}}[/tex]
⇛[tex]\huge\tt\frac{ \sqrt{tanx} }{sinxcosx}[/tex]
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⇛[tex]\huge\tt \frac{ \sqrt{tanx} }{sinxcosx \times \frac{cosx}{cosx}}[/tex]
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⇛[tex]\huge\tt \frac{ \sqrt{tanx} }{sinx \times \frac{ {cos}^{2} x}{cosx}}[/tex] ㅤ ㅤ ㅤ
⇛ [tex]\huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2} x \times \frac{sinx}{cosx} }[/tex]
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⇛[tex]\huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2}x \times tanx }[/tex]
⇛[tex]\huge\tt {tan}^{ \frac{1}{2} – 1 } \times \frac{1}{ {cos}^{2} x}[/tex]ㅤ ㅤ ㅤ ㅤ ㅤ
⇛[tex]\huge\tt {(tan)}^{ – \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = {(tanx)}^{ – \frac{1}{2} } \times {sec}^{2} x⇛(tan)[/tex]
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⇛[tex]\huge\tt {(tan)}^{ – \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = ∫ {(tanx)}^{ – \frac{1}{2} } \times {sec}^{2} x \times dx⇛(tan)[/tex]
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[tex]\bold\blue{☛\: Let tanx=t}[/tex]
[tex]\bold\blue{☛ \:Differentiating \: both \: sides \: w.r.t.x}[/tex]
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⇛[tex]\huge\tt {sec}^{2} x = \frac{dt}{dx}[/tex]
⇛[tex]\huge\tt{dx \frac{dt}{ {sec}^{2}x }}[/tex]
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⇛[tex]\huge\tt∴∫ {(tanx)}^{ – \frac{1}{2} } \times {sec}^{2} x \times dx[/tex]
⇛[tex]\huge\tt ∫ {(t)}^{ – \frac{1}{2} } \times {sec}^{2} x \times \frac{dt}{ {sec}^{2}x }[/tex]
⇛[tex]\huge\tt ∫ {t}^{ – \frac{1}{2} }[/tex]ㅤ ㅤ
⇛ [tex]\huge\tt\frac{ {t}^{ – \frac{1}{2} + 1} }{ – \frac{1}{2} + 1 }[/tex]
⇛ [tex]\huge\tt \frac{ {t}^{ \frac{1}{2} } }{ \frac{1}{2} } + c = 2 {t}^{ \frac{1}{2} } + c = 2 \sqrt{t}[/tex]
⇛[tex]\huge2 \sqrt{t} + c = 2 \sqrt{tanx}[/tex]
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Answer:
[tex]⇛\huge2 \sqrt{t} + c = 2 \sqrt{tanx}2t+c=2tanx
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[/tex]
is the answer. hope it helps you. thanks.