Answer :
Solution: [tex] \int\limits \: \sqrt{tan(x)} \: dx[/tex]
1) pass: make the replacement
[tex]y = \sqrt{tan(x)} = tan(x)^ \frac{1}{2} [/tex]
[tex]dy = \frac{1}{2} *(tan\:x)^{- \frac{1}{2}} * sec^2x\:dx[/tex]
[tex]2dy = \frac{1}{y}*(tan^2*x+1)dx[/tex]
[tex]2ydy = (y^4+1)dx[/tex]
[tex]dx = \frac{2y}{y^4+1} dy[/tex]
2) pass: We substitute in the integral of the statement
[tex]I = \int\limits \sqrt{tanx} \: dx[/tex]
[tex]I = \int\limits \:y* \frac{2y}{y^4+1} dy[/tex]
[tex]I = \int\limits\: \frac{2y^2}{y^4+1} dy[/tex]
3) pass: Using the Gauss Lemma, we will factor the polynomial (y⁴ + 1) knowing that there are no real roots, so we will directly try to factorize into two polynomials of degree 2:
[tex]y^4+1 = (y^2+ay+1)(y^2+cy+1)[/tex]
[tex]y^4+1 = y^4+(a+c)y^3 + (ac+2)y^2+(a+c)y+1[/tex]
Make the system linear, to find the values of "a" and "c".
[tex] \left \{ {{a+c=0} \atop {ac+2=0}} \right. [/tex]
[tex]a+c = 0 \:\to c=-a[/tex]
[tex]ac+2=0\to ac = -2\to a*(-a) = -2\to -a^2 = -2\to a =\sqrt{2}[/tex]
[tex]a+c = 0 \:\to c=-a\to c = - \sqrt{2} [/tex]
We have:
[tex](y^4+1)= (y^2+ \sqrt{2y} +1)(y^2- \sqrt{2y} +1)[/tex]
4) pass: We will use the partial fractions method, using the fraction from within the integral:
[tex] \frac{2y^2}{y^4+1} = \frac{Ay+B}{y^2+ \sqrt{2y}+1 } + \frac{Cy+D}{y^2- \sqrt{2y} +1} [/tex]
[tex]= \frac{(A+C)y^3+(- \sqrt{2}A+B+ \sqrt{2}C+D)y^2+(A- \sqrt{2}B+C+ \sqrt{2}D)y+(B+D) }{y^4+1} [/tex]
[tex]\longrightarrow \left \{ {{A+C=0\to A=-C} \atop { -\sqrt{2}(A-C)+(B+D)=2 }} \right. [/tex]
[tex]-\sqrt{2}(A-C)+(B+D)=2 \to - \sqrt{2} *2A+0=2\to A = - \frac{1}{ \sqrt{2} } [/tex]
[tex]A+C=0\to C=-A\to C = - (- \frac{1}{ \sqrt{2} } )\to C = \frac{1}{ \sqrt{2} } [/tex]
[tex](A+C)+ \sqrt{2} (D-B)=0\to 0+ \sqrt{2} (D-B)=0 \to B=D=0[/tex]
[tex]B+D=0[/tex]
5)pass: Adopt what was used above:
[tex]I = \int\limits \frac{ -\frac{1}{ \sqrt{2} }y }{y^2+ \sqrt{2}y+1 } dy+ \int\limits \frac{ \frac{1}{ \sqrt{2} }y }{y^2- \sqrt{2}y+1 } dy[/tex]
[tex]I = - \frac{1}{ \sqrt{2} } \underbrace{\int\limits \frac{y}{y^2+ \sqrt{2y}+1 }dy }_{I_1}+ \frac{1}{ \sqrt{2} } \underbrace{\int\limits \frac{y}{y^2- \sqrt{2y}+1 }dy }_{I_2}[/tex]
6) pass: Now, solve it separately
[tex]I_1 = \int\limits \frac{y}{y^2+ \sqrt{2y}+1 }dy[/tex]
[tex]2I_1 = \int\limits\frac{2y}{y^2+ \sqrt{2y}+1 }dy = \int\limits \frac{2y- \sqrt{2}+ \sqrt{2} }{y^2+ \sqrt{2y}+1 } dy[/tex]
[tex]2I_1 = \int\limits\frac{2y+ \sqrt{2} }{y^2+ \sqrt{2y}+1 }dy - \int\limits\frac{ \sqrt{2} }{y^2+ \sqrt{2y}+1 }dy[/tex]
[tex]2I_1 = ln|y^2+ \sqrt{2}y+1| - \sqrt{2} \int\limits \frac{1}{(y \sqrt{2}+1)^2+1 }dy [/tex]
[tex]\boxed{I_1 = \frac{1}{2} ln|y^2+ \sqrt{2} y+1| - \frac{ \sqrt{2} }{2} arctan(y \sqrt{2}+1 )}[/tex]
and
[tex]I_2 = \int\limits \frac{y}{y^2- \sqrt{2y}+1 }dy[/tex]
[tex]2I_2 = \int\limits\frac{2y}{y^2- \sqrt{2y}+1 }dy = \int\limits \frac{2y- \sqrt{2}+ \sqrt{2} }{y^2+ \sqrt{2y}+1 } dy[/tex]
[tex]2I_2 = \int\limits\frac{2y- \sqrt{2} }{y^2- \sqrt{2y}+1 }dy - \int\limits\frac{ \sqrt{2} }{y^2- \sqrt{2y}+1 }dy[/tex]
[tex]2I_2 = ln|y^2- \sqrt{2}y+1| + \sqrt{2} \int\limits \frac{1}{(y \sqrt{2}-1)^2+1 }dy [/tex]
[tex]\boxed{I_2 = \frac{1}{2} ln|y^2- \sqrt{2} y+1| + \frac{ \sqrt{2} }{2} arctan(y \sqrt{2}-1 )}[/tex]
7) pass: Let's use the expression of [tex]I[/tex]
[tex]I = - \frac{I_1}{ \sqrt{2} } + \frac{I_2}{ \sqrt{2} } [/tex]
[tex]I = - \frac{ \frac{1}{2}ln|y^2+ \sqrt{2}y+1|- \frac{ \sqrt{2} }{2}arctan(y \sqrt{2} + 1) }{ \sqrt{2} } + \frac{ \frac{1}{2}ln|y^2- \sqrt{2}y+1|+ \frac{ \sqrt{2} }{2}arctan(y \sqrt{2} - 1) }{ \sqrt{2} }[/tex]
[tex]I = - \frac{1}{2 \sqrt{2} } ln|y^2+ \sqrt{2} y +1|+\frac{1}{ \sqrt{2} } arctan(y \sqrt{2} +1)+[/tex]
[tex]+\frac{1}{2 \sqrt{2} } ln|y^2- \sqrt{2} y +1|+\frac{1}{ \sqrt{2} } arctan(y \sqrt{2}-1)[/tex]
8) pass: Now, returning to the expression as a function of x, we finally arrive at the final answer:
[tex]I = - \frac{1}{2 \sqrt{2} } ln|y^2+ \sqrt{2} y +1|+\frac{1}{ \sqrt{2} } arctan(y \sqrt{2} +1)+[/tex]
[tex]+\frac{1}{2 \sqrt{2} } ln|y^2- \sqrt{2} y +1|+\frac{1}{ \sqrt{2} } arctan(y \sqrt{2}-1)[/tex]
[tex]I = - \frac{1}{2 \sqrt{2} } ln|( \sqrt{tan\:x})^2+ \sqrt{2}( \sqrt{tan\:x}) +1| + \frac{1}{ \sqrt{2} } arctan(( \sqrt{tan\:x}) \sqrt{2} +1[/tex]
[tex]+\frac{1}{2 \sqrt{2} } ln|( \sqrt{tan\:x})^2- \sqrt{2}( \sqrt{tan\:x}) +1| + \frac{1}{ \sqrt{2} } arctan(( \sqrt{tan\:x}) \sqrt{2} -1[/tex]
Answer:
[tex]\boxed{I = - \frac{1}{2 \sqrt{2} } ln | ( \ tan\:x+\sqrt{2\:tan\:x} +1| + \frac{1}{ \sqrt{2} }arctan( \sqrt{2\:tan\:x} +1) +}[/tex]
[tex]\boxed{+ \frac{1}{2 \sqrt{2} } ln | ( \ tan\:x-\sqrt{2\:tan\:x} +1| + \frac{1}{ \sqrt{2} }arctan( \sqrt{2\:tan\:x} -1) +C}}[/tex] [tex]\end{array}}\qquad\quad\checkmark[/tex]
1) pass: make the replacement
[tex]y = \sqrt{tan(x)} = tan(x)^ \frac{1}{2} [/tex]
[tex]dy = \frac{1}{2} *(tan\:x)^{- \frac{1}{2}} * sec^2x\:dx[/tex]
[tex]2dy = \frac{1}{y}*(tan^2*x+1)dx[/tex]
[tex]2ydy = (y^4+1)dx[/tex]
[tex]dx = \frac{2y}{y^4+1} dy[/tex]
2) pass: We substitute in the integral of the statement
[tex]I = \int\limits \sqrt{tanx} \: dx[/tex]
[tex]I = \int\limits \:y* \frac{2y}{y^4+1} dy[/tex]
[tex]I = \int\limits\: \frac{2y^2}{y^4+1} dy[/tex]
3) pass: Using the Gauss Lemma, we will factor the polynomial (y⁴ + 1) knowing that there are no real roots, so we will directly try to factorize into two polynomials of degree 2:
[tex]y^4+1 = (y^2+ay+1)(y^2+cy+1)[/tex]
[tex]y^4+1 = y^4+(a+c)y^3 + (ac+2)y^2+(a+c)y+1[/tex]
Make the system linear, to find the values of "a" and "c".
[tex] \left \{ {{a+c=0} \atop {ac+2=0}} \right. [/tex]
[tex]a+c = 0 \:\to c=-a[/tex]
[tex]ac+2=0\to ac = -2\to a*(-a) = -2\to -a^2 = -2\to a =\sqrt{2}[/tex]
[tex]a+c = 0 \:\to c=-a\to c = - \sqrt{2} [/tex]
We have:
[tex](y^4+1)= (y^2+ \sqrt{2y} +1)(y^2- \sqrt{2y} +1)[/tex]
4) pass: We will use the partial fractions method, using the fraction from within the integral:
[tex] \frac{2y^2}{y^4+1} = \frac{Ay+B}{y^2+ \sqrt{2y}+1 } + \frac{Cy+D}{y^2- \sqrt{2y} +1} [/tex]
[tex]= \frac{(A+C)y^3+(- \sqrt{2}A+B+ \sqrt{2}C+D)y^2+(A- \sqrt{2}B+C+ \sqrt{2}D)y+(B+D) }{y^4+1} [/tex]
[tex]\longrightarrow \left \{ {{A+C=0\to A=-C} \atop { -\sqrt{2}(A-C)+(B+D)=2 }} \right. [/tex]
[tex]-\sqrt{2}(A-C)+(B+D)=2 \to - \sqrt{2} *2A+0=2\to A = - \frac{1}{ \sqrt{2} } [/tex]
[tex]A+C=0\to C=-A\to C = - (- \frac{1}{ \sqrt{2} } )\to C = \frac{1}{ \sqrt{2} } [/tex]
[tex](A+C)+ \sqrt{2} (D-B)=0\to 0+ \sqrt{2} (D-B)=0 \to B=D=0[/tex]
[tex]B+D=0[/tex]
5)pass: Adopt what was used above:
[tex]I = \int\limits \frac{ -\frac{1}{ \sqrt{2} }y }{y^2+ \sqrt{2}y+1 } dy+ \int\limits \frac{ \frac{1}{ \sqrt{2} }y }{y^2- \sqrt{2}y+1 } dy[/tex]
[tex]I = - \frac{1}{ \sqrt{2} } \underbrace{\int\limits \frac{y}{y^2+ \sqrt{2y}+1 }dy }_{I_1}+ \frac{1}{ \sqrt{2} } \underbrace{\int\limits \frac{y}{y^2- \sqrt{2y}+1 }dy }_{I_2}[/tex]
6) pass: Now, solve it separately
[tex]I_1 = \int\limits \frac{y}{y^2+ \sqrt{2y}+1 }dy[/tex]
[tex]2I_1 = \int\limits\frac{2y}{y^2+ \sqrt{2y}+1 }dy = \int\limits \frac{2y- \sqrt{2}+ \sqrt{2} }{y^2+ \sqrt{2y}+1 } dy[/tex]
[tex]2I_1 = \int\limits\frac{2y+ \sqrt{2} }{y^2+ \sqrt{2y}+1 }dy - \int\limits\frac{ \sqrt{2} }{y^2+ \sqrt{2y}+1 }dy[/tex]
[tex]2I_1 = ln|y^2+ \sqrt{2}y+1| - \sqrt{2} \int\limits \frac{1}{(y \sqrt{2}+1)^2+1 }dy [/tex]
[tex]\boxed{I_1 = \frac{1}{2} ln|y^2+ \sqrt{2} y+1| - \frac{ \sqrt{2} }{2} arctan(y \sqrt{2}+1 )}[/tex]
and
[tex]I_2 = \int\limits \frac{y}{y^2- \sqrt{2y}+1 }dy[/tex]
[tex]2I_2 = \int\limits\frac{2y}{y^2- \sqrt{2y}+1 }dy = \int\limits \frac{2y- \sqrt{2}+ \sqrt{2} }{y^2+ \sqrt{2y}+1 } dy[/tex]
[tex]2I_2 = \int\limits\frac{2y- \sqrt{2} }{y^2- \sqrt{2y}+1 }dy - \int\limits\frac{ \sqrt{2} }{y^2- \sqrt{2y}+1 }dy[/tex]
[tex]2I_2 = ln|y^2- \sqrt{2}y+1| + \sqrt{2} \int\limits \frac{1}{(y \sqrt{2}-1)^2+1 }dy [/tex]
[tex]\boxed{I_2 = \frac{1}{2} ln|y^2- \sqrt{2} y+1| + \frac{ \sqrt{2} }{2} arctan(y \sqrt{2}-1 )}[/tex]
7) pass: Let's use the expression of [tex]I[/tex]
[tex]I = - \frac{I_1}{ \sqrt{2} } + \frac{I_2}{ \sqrt{2} } [/tex]
[tex]I = - \frac{ \frac{1}{2}ln|y^2+ \sqrt{2}y+1|- \frac{ \sqrt{2} }{2}arctan(y \sqrt{2} + 1) }{ \sqrt{2} } + \frac{ \frac{1}{2}ln|y^2- \sqrt{2}y+1|+ \frac{ \sqrt{2} }{2}arctan(y \sqrt{2} - 1) }{ \sqrt{2} }[/tex]
[tex]I = - \frac{1}{2 \sqrt{2} } ln|y^2+ \sqrt{2} y +1|+\frac{1}{ \sqrt{2} } arctan(y \sqrt{2} +1)+[/tex]
[tex]+\frac{1}{2 \sqrt{2} } ln|y^2- \sqrt{2} y +1|+\frac{1}{ \sqrt{2} } arctan(y \sqrt{2}-1)[/tex]
8) pass: Now, returning to the expression as a function of x, we finally arrive at the final answer:
[tex]I = - \frac{1}{2 \sqrt{2} } ln|y^2+ \sqrt{2} y +1|+\frac{1}{ \sqrt{2} } arctan(y \sqrt{2} +1)+[/tex]
[tex]+\frac{1}{2 \sqrt{2} } ln|y^2- \sqrt{2} y +1|+\frac{1}{ \sqrt{2} } arctan(y \sqrt{2}-1)[/tex]
[tex]I = - \frac{1}{2 \sqrt{2} } ln|( \sqrt{tan\:x})^2+ \sqrt{2}( \sqrt{tan\:x}) +1| + \frac{1}{ \sqrt{2} } arctan(( \sqrt{tan\:x}) \sqrt{2} +1[/tex]
[tex]+\frac{1}{2 \sqrt{2} } ln|( \sqrt{tan\:x})^2- \sqrt{2}( \sqrt{tan\:x}) +1| + \frac{1}{ \sqrt{2} } arctan(( \sqrt{tan\:x}) \sqrt{2} -1[/tex]
Answer:
[tex]\boxed{I = - \frac{1}{2 \sqrt{2} } ln | ( \ tan\:x+\sqrt{2\:tan\:x} +1| + \frac{1}{ \sqrt{2} }arctan( \sqrt{2\:tan\:x} +1) +}[/tex]
[tex]\boxed{+ \frac{1}{2 \sqrt{2} } ln | ( \ tan\:x-\sqrt{2\:tan\:x} +1| + \frac{1}{ \sqrt{2} }arctan( \sqrt{2\:tan\:x} -1) +C}}[/tex] [tex]\end{array}}\qquad\quad\checkmark[/tex]