Question

help me ∫{√tanx +√cotx}dx

Answer 1

write tanx as sinx/cosx and cotx= cosx/sinx.. then it will convert into ∫{ (sinx + cosx) /√sinxcox} .. then multiply and divide it by √2.. it will convert into ∫{√2 (sinx + cosx)dx/√sin2x} ... then put (sinx–cosx) =t ..& for finding sin2x.. square both side u will get the value of sin2x in terms of "t" .. it will reduce to ∫√2{dt/(√1–t²)} ..
now u can do it on own.. hope it helped.. mark as brainliest

Answer 2

∫{√tanx +√cotx}dx
= ∫{1/√cotx +√cotx}dx
= ∫(cotx+1/√cotx)dx
= ∫{√tanx(cotx+1)}dx

let tanx = t²
=> sec²x = 2tdt

1+tan²x = 2tdt/dx
1+(t²)² = 2t.dt/dx
1+t⁴ = 2tdt/dx
=> dx = ( 2t/1+t⁴ )dt

using this we get

= ∫t[1/t²+1] × 2t/1+t² .dt
= ∫t{1+t²/t²} × 2t/1+t² .dt
= ∫2[1+t²/1+t⁴] dt
= 2 ∫ 1+t²/1+t⁴ dt

Dividing and multiplying by t²
=> 2 ∫(1/t²+1)/(1/t²+t²) . dt
adding and subtracting 2 in the denominator
= 2 ∫(1/t²+1)/(1/t²+t²+2-2 ) . dt
=>2 ∫(1/t²+1)/(t-1/t)²+(√2)² {©ompleting square
Now let t-1/t = y
=> 1+1/t² dt = dy
=> dt = dy/(1+1/t²)

Now substituting these values we get :-
= 2 ∫dy/t²+(√2)².dy
=> 2{ 1/√2 tan-¹y/√2 + C}

= √2 tan-¹ 1/t-t/√2 + C { y = 1/t-t
=> √2 tan-¹ (t²-1/√2t) +C
= √2 tan-¹(tanx-1/√2tanx) +C