intigration of sqrt(tanx)

Question

calculs of Dr. M. Yusuf,

Answer 1

Let y²=tan(x), 2ydy=sec²(x)dx=(1+tan²(x))dx=(1+y⁴)dx, dx=2ydy/(1+y⁴).

Let J=∫√tan(x)dx=2∫y²dy/(1+y⁴)=2∫dy/(y²+1/y²).

y²+1/y²=(y+1/y)²-2 or (y-1/y)²+2. Let p=y+1/y and q=y-1/y, so:

dp=dy(1-1/y²) and dq=dy(1+1/y²), hence dp+dq=2dy=dy(1-1/y²)+dy(1+1/y²).

And 2dy/(y²+1/y²)=dy(1-1/y²)/(y²+1/y²)+dy(1+1/y²)/(y²+1/y²)=dp/(p²-2)+dq/(q²+2).

J=∫dp/(p²-2)+∫dq/(q²+2).

HYPERBOLIC FUNCTIONS (sinh, cosh, tanh, sech and some derivatives)

Consider cosh²(z)-sinh²(z)=1 where cosh(z)=½(e^z+e^-z) and sinh(z)=½(e^z-e^-z), and tanh(z)=sinh(z)/cosh(z)=(e^z-e^-z)/(e^z-e^-z).

Also d(sinh(z))=cosh(x), d(cosh(z))=sinh(z), d(tanh(z))=(cosh²(z)-sinh²(z))/cosh²(z)=1/cosh²(z)=sech²(z).

1-tanh²(z)=sech²(z), or tanh²(z)+sech²(z)=1.

INTEGRATE BY SUBSTITUTION

∫dp/(p²-2): Let p=tanh(u)√2, dp=sech²(u)√2du; p²-2=2tanh²(u)-2=-2sech²(u).

∫dp/(p²-2)=-½∫sech²(u)√2du/sech²(u)=-(√2/2)∫du=-(√2/2)u=-(√2/2)tanh^-1(p/√2), or -(√2/2)tanh^-1(p√2/2).

-(√2/2)tanh^-1(p√2/2)=-(√2/2)tanh^-1(√2/2(√tan(x)+√cot(x))).

∫dq/(q²+2)=(√2/2)tan^-1(q√2/2)=(√2/2)tan^-1(√tan(x)-√cot(x)).

J=-(√2/2)tanh^-1(√2/2(√tan(x)+√cot(x)))+(√2/2)tan^-1(√tan(x)-√cot(x))+C where C is integration constant.

(To integrate dq/(q²+2), from first principles, the substitution is q=tan(u)√2, etc., similar to above for tanh.)