Question

Integrate:-


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

Answer 1

Answer:

$_ _ _ _ _ _ _ _ _ _ _ _ _ _✍️</p><p></p><p>⇛\huge\tt\frac{ \sqrt{tanx} }{sinxcosx}sinxcosxtanx</p><p></p><p>ㅤ ㅤ ㅤ ㅤ ㅤ</p><p></p><p>⇛\huge\tt \frac{ \sqrt{tanx} }{sinxcosx \times \frac{cosx}{cosx}}sinxcosx×cosxcosxtanx</p><p></p><p>ㅤ ㅤ ㅤ ㅤ ㅤ</p><p></p><p>⇛\huge\tt \frac{ \sqrt{tanx} }{sinx \times \frac{ {cos}^{2} x}{cosx}}sinx×cosxcos2xtanx ㅤ ㅤ ㅤ</p><p></p><p>⇛ \huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2} x \times \frac{sinx}{cosx} }cos2x×cosxsinxtanx</p><p></p><p>ㅤ ㅤ ㅤ ㅤ ㅤ</p><p></p><p>⇛\huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2}x \times tanx }cos2x×tanxtanx</p><p></p><p>⇛\huge\tt {tan}^{ \frac{1}{2} - 1 } \times \frac{1}{ {cos}^{2} x}tan21−1×cos2x1 ㅤ ㅤ ㅤ ㅤ ㅤ</p><p></p><p>⇛\huge\tt {(tan)}^{ - \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x⇛(tan)(tan)−21×cos2x1=(tanx)−21×sec2x⇛(tan)</p><p></p><p>ㅤ ㅤ ㅤ ㅤ ㅤ</p><p></p><p>⇛\huge\tt {(tan)}^{ - \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = ∫ {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x \times dx⇛(tan)(tan)−21×cos2x1=∫(tanx)−21×sec2x×dx⇛(tan)</p><p></p><p>ㅤ ㅤ ㅤ ㅤ ㅤ</p><p></p><p>\bold\blue{☛\: Let tanx=t}☛Lettanx=t</p><p></p><p>\bold\blue{☛ \:Differentiating \: both \: sides \: w.r.t.x}☛Differentiatingbothsidesw.r.t.x</p><p></p><p>ㅤ ㅤ ㅤ ㅤ ㅤ</p><p></p><p>⇛\huge\tt {sec}^{2} x = \frac{dt}{dx}sec2x=dxdt</p><p></p><p>⇛\huge\tt{dx \frac{dt}{ {sec}^{2}x }}dxsec2xdt</p><p></p><p>ㅤ ㅤ ㅤ ㅤ ㅤ</p><p></p><p>⇛\huge\tt∴∫ {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x \times dx∴∫(tanx)−21×sec2x×dx</p><p></p><p>⇛\huge\tt ∫ {(t)}^{ - \frac{1}{2} } \times {sec}^{2} x \times \frac{dt}{ {sec}^{2}x }∫(t)−21×sec2x×sec2xdt</p><p></p><p>⇛\huge\tt ∫ {t}^{ - \frac{1}{2} }∫t−21 ㅤ ㅤ</p><p></p><p>⇛ \huge\tt\frac{ {t}^{ - \frac{1}{2} + 1} }{ - \frac{1}{2} + 1 }−21+1t−21+1</p><p></p><p>⇛ \huge\tt \frac{ {t}^{ \frac{1}{2} } }{ \frac{1}{2} } + c = 2 {t}^{ \frac{1}{2} } + c = 2 \sqrt{t}21t21+c=2t21+c=2t</p><p></p><p>⇛\huge2 \sqrt{t} + c = 2 \sqrt{tanx}2t+c=2tanx</p><p></p><p>$

Step-by-step explanation:

mark as brainlist answer

Answer 2

Answer:

Hello mate, answer is here!!

Step-by-step explanation:

I = ( 1 + sinx)dx /( sinx + sinx .cosx)

we Know ,

sin2x = 2tanx/( 1+ tan²x)

cos2x = (1 - tan²x)/( 1 + tan²x)

use this

I = { 1 + 2tanx/2/( 1+ tan²x/2)}dx/{ 2tanx/2/( 1+tan²/2.) + 2tanx/2( 1 - tan²x/2)/( 1+ tan²x/2)² }

= ( 1 + tan²x/2 +2tanx/2)dx/( 2tanx/2)( 1+ tan²x/2 + 1 - tan²x/2)/( 1 + tan²x/2)

= ( 1 + tan²x/2)( 1 + tanx/2)²dx/2tanx/2× 2

= (1/4) { (sec²x/2)( 1 +tanx/2)dx/tanx/2

[ note :- ( 1 + tan²x/2 ) = sec²x/2 ]

now , let tanx/2 = P

differentiate

sec²x/2 × 1/2 dx = dP

sec²x/2dx = 2dP

use this in above

I = ( 1 + P)²2.dP/4P

=1/2 × ( 1+P² + 2P)dP/P

=1/2 × { dP/P + P.dP + 2dP }

= 1/2× { lnP + P²/2 + 2P }

put P = tanx/2

I = 1/2tanx/2 + 1/4tan²x/2 + tanx + C.