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4v2 ( sin 3x + sin x) has value equal to : Lt *2 2 sin 2x sin+ 1 5X + cos -V2 + 2 cos 2x + cos 3X |
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Answer» ONG>Answer:
Integrate the function \huge\green\tt\FRAC{ \sqrt{tanx} }{SINXCOSX}} ⇛\huge\tt\frac{ \sqrt{tanx} }{sinxcosx} sinxcosx tanx
ㅤ ㅤ ㅤ ㅤ ㅤ ⇛\huge\tt \frac{ \sqrt{tanx} }{sinxcosx \times \frac{cosx}{cosx}} sinxcosx× cosx cosx
tanx
ㅤ ㅤ ㅤ ㅤ ㅤ ⇛\huge\tt \frac{ \sqrt{tanx} }{sinx \times \frac{ {cos}^{2} x}{cosx}} sinx× cosx cos 2 x
tanx
ㅤ ㅤ ㅤ ⇛ \huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2} x \times \frac{sinx}{cosx} } cos 2 x× cosx sinx
tanx
ㅤ ㅤ ㅤ ㅤ ㅤ ⇛\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
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)(tan) − 2 1
× cos 2 x 1
=(tanx) − 2 1
×sec 2 x⇛(tan) 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 \times DX⇛(tan)(tan) − 2 1
× cos 2 x 1
=∫(tanx) − 2 1
×sec 2 x×dx⇛(tan) ㅤ ㅤ ㅤ ㅤ ㅤ \bold\blue{☛\: Let tanx=t}☛Lettanx=t \bold\blue{☛ \:Differentiating \: both \: sides \: w.r.t.x}☛Differentiatingbothsidesw.r.t.x ㅤ ㅤ ㅤ ㅤ ㅤ ⇛\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
ㅤ ㅤ ㅤ ㅤ ㅤ ⇛\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
ㅤ ㅤ ⇛ \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
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