find the intergarl of root tanx
hello abhishek
finding integral of root tanx
∫√(tan x) dx
Let tan x = t
2
⇒ sec
2
x dx = 2t dt
⇒ dx = [2t / (1 + t
4
)]dt
⇒ Integral ∫ 2t
2
/ (1 + t
4
) dt
⇒ ∫[(t
2
+ 1) + (t
2
- 1)] / (1 + t
4
) dt
⇒ ∫(t
2
+ 1) / (1 + t
4
) dt + ∫(t
2
- 1) / (1 + t
4
) dt
⇒ ∫(1 + 1/t
2
) / (t
2
+ 1/t
2
) dt + ∫(1 - 1/t
2
) / (t
2
+ 1/t
2
) dt
⇒ ∫(1 + 1/t
2
)dt / [(t - 1/t)
2
+ 2] + ∫(1 - 1/t
2
)dt / [(t + 1/t)
2
-2]
Let t - 1/t = u for the first integral ⇒ (1 + 1/t
2
)dt = du
and t + 1/t = v for the 2nd integral ⇒ (1 - 1/t
2
)dt = dv
Integral
= ∫du/(u
2
+ 2) + ∫dv/(v
2
- 2)
= (1/√2) tan
-1
(u/√2) + (1/2√2) log(v -√2)/(v + √2)l + c
= (1/√2) tan
-1
[(t
2
- 1)/t√2] + (1/2√2) log (t
2
+ 1 - t√2) / t
2
+ 1 + t√2) + c
= (1/√2) tan
-1
[(tanx - 1)/(√2tan x)] + (1/2√2) log [tanx + 1 - √(2tan x)] / [tan x + 1 + √(2tan x)] + c
if you still are unable to understand the method, you can also refer to this video: https://www.youtube.com/watch?v=CcP7YKr5qBo
hope this helps