131/2[∫√tanx-√cotx dx +∫√tanx+√cotx dx]
9 Answers
13first one is a tricky step ...then rewrite tan and cot in sin and cos
i m too lazy
13otherwise
v = sqrt(tan(x))
dv = sec2(x) / (2 sqrt(tan(x)))
dv = 1 / (2 cos2(x) sqrt(tan(x)))
∫sqrt(tan(x)) dx
∫sqrt(tan(x)) dx/ (cos2(x) + sin2(x))
∫sqrt(tan(x)) dx/ (cos2(x) (1+tan2(x)))
∫tan(x) dx/ (cos2(x) sqrt(tan(x)) (1+tan2(x)))
∫2tan(x) dx/ (2 cos2(x) sqrt(tan(x)) (1+tan2(x)))
∫2v2 dv / (1+v4)
1+v4 = (v2-sqrt(2)v+1)(v2+sqrt(2)v+1)
So the same integral can now be expressed using this identity as
∫2v2 dv / ((v2-sqrt(2)v+1)(v2+sqrt(2)v+1))
And now, breaking it down into partial fractions, the integral becomes,
∫(sqrt(2)/2) v dv / (v2-sqrt(2)v+1) + (-sqrt(2)/2) v dv / (v2+sqrt(2)v+1)
Now this is starting to get tricky, so I'll break the first term into two integrals I1 and I2, and the second term into two more integrals I3 and I4.
∫sqrt(tan(x)) dx is I1 + I2 + I3 + I4, where
I1 = ∫(sqrt(2)/4)(2v-sqrt(2)) dv / (v2-sqrt(2)v+1)
I2 = ∫(1/2) dv / (v2-sqrt(2)v+1)
I3 = ∫(-sqrt(2)/4)(2v+sqrt(2)) dv / (v2+sqrt(2)v+1)
I4 = ∫(1/2) dv / (v2+sqrt(2)v+1)
Now I'll do each of I1, I2, I3, and I4 separately:
I1 = ∫(sqrt(2)/4)(2v-sqrt(2)) dv / (v2-sqrt(2)v+1)
This integral is of the form ∫du/u, which is ln|u|, so
I1 = (sqrt(2)/4) ln|v2-sqrt(2)v+1| + C1
I2 = ∫(1/2) dv / (v2-sqrt(2)v+1)
This can be converted into the form a/((av+b)2+1), if we let a=sqrt(2) and b=-1
I2 = ∫(sqrt(2)/2) sqrt(2)/(2v2-2sqrt(2)v+1+1)
I2 = ∫(sqrt(2)/2) sqrt(2)/((sqrt(2)v-1)2+1)
I2 = (sqrt(2)/2) atan(sqrt(2)v-1) + C2
I3 = ∫(-sqrt(2)/4)(2v+sqrt(2)) dv / (v2+sqrt(2)v+1)
Again, this is the ∫the ln form, so
I3 = (-sqrt(2)/4) ln(v2+sqrt(2)v+1) + C3
I4 = ∫(1/2) dv / (v2+sqrt(2)v+1)
Again, this can be converted to the atan form, so
I4 = (sqrt(2)/2) atan(sqrt(2)v+1) + C4
To summarize,
I1 = (sqrt(2)/4) ln(v2-sqrt(2)v+1) + C1
I2 = (sqrt(2)/2) atan(sqrt(2)v-1) + C2
I3 = (-sqrt(2)/4) ln(v2+sqrt(2)v+1) + C3
I4 = (sqrt(2)/2) atan(sqrt(2)v+1) + C4
The sum of which gives us the final answer,
∫sqrt(tan(x)) =
(sqrt(2)/4) ln(v2-sqrt(2)v+1) + (sqrt(2)/2) atan(sqrt(2)v-1) + (-sqrt(2)/4) ln(v2+sqrt(2)v+1) + (sqrt(2)/2) atan(sqrt(2)v+1) + C =
(sqrt(2)/4) ln(tan(x)-sqrt(2tan(x))+1) + (sqrt(2)/2) atan(sqrt(2tan(x))-1) +
(-sqrt(2)/4) ln(tan(x)+sqrt(2tan(x))+1) + (sqrt(2)/2) atan(sqrt(2tan(x))+1) + C
copied from other source.....
11My most favourite question in integration. Another method:
Put tanx=t2
x=tan-1x2
dx=2tdt/1+t4
Question becomes........∫t.2tdt/1+t4
=∫2t2dt/1+t4
=∫(t2+1+t2-1)dt/1+t4
=∫(t2+1)dt/1+t4 + ∫(t2-1)dt/1+t4
Let first part be I1 and second part be I2.
I1=∫(t2+1)dt/1+t4
Divide Nr and Dr by t2
11It becomes
I1= ∫(1+1/t2)dt/(t2+1/t2)
And now the master piece........
Let t-1/t = u
Then (1+1/t2)dt = du
t2+1/t2 = u2+2 (verify yourselves)
Thus I1=∫du/u2+2
= 1/√2 tan-1(u/√2)
For I2, divide both Nr and Dr by t2 and let 1+1/t = v
Then dv=(1-1/t2) dt
Again, t2+1/t2 = v2-2
I2=∫dv/v2-2
=(1/2√2)logl(v-√2)/(v+√2)l
Combine I1 and I2 and substitute for u and v by t and again substitute for t as √tanx
HOPE U UNDERSTAND [1]
13my favourite method
I=1/2[∫√tanx-√cotx dx +∫√tanx+√cotx dx]
=1/2[∫√2{√sinx +√cosx}/√1-(sinx-cosx)2 dx +∫1/2[√2∫{√sinx -√cosx}/√(sinx+cosx)2-1 dx]
=1/√2[∫dt/√1-t2-∫dm/√m2-1]
[taking sinx-cosx=p and sinx+cosx=m]
now it is simple.....
