Math question! \int\ { √(tan(x)) } \, dx Please show your work!

Question

Math question!


`\int\ { √(tan(x)) } \, dx`

Please show your work!

Answer 1

Solution:
`\int\limits \: √(tan(x)) \: dx`
1) pass: make the replacement

`y = √(tan(x)) = tan(x)^ (1)/(2)`

`dy = (1)/(2) *(tan\:x)^{- (1)/(2)} * sec^2x\:dx`

`2dy = (1)/(y)*(tan^2*x+1)dx`

`2ydy = (y^4+1)dx`

`dx = (2y)/(y^4+1) dy`

2) pass: We substitute in the integral of the statement

`I = \int\limits √(tanx) \: dx`

`I = \int\limits \:y* (2y)/(y^4+1) dy`

`I = \int\limits\: (2y^2)/(y^4+1) dy`

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:


`y^4+1 = (y^2+ay+1)(y^2+cy+1)`

`y^4+1 = y^4+(a+c)y^3 + (ac+2)y^2+(a+c)y+1`
Make the system linear, to find the values ​​of "a" and "c".

`\left \{ {{a+c=0} \atop {ac+2=0}} \right.`

`a+c = 0 \:\to c=-a`

`ac+2=0\to ac = -2\to a*(-a) = -2\to -a^2 = -2\to a =√(2)`

`a+c = 0 \:\to c=-a\to c = - √(2)`
We have:

`(y^4+1)= (y^2+ √(2y) +1)(y^2- √(2y) +1)`

4) pass: We will use the partial fractions method, using the fraction from within the integral:

`(2y^2)/(y^4+1) = (Ay+B)/(y^2+ √(2y)+1 ) + (Cy+D)/(y^2- √(2y) +1)`

`= ((A+C)y^3+(- √(2)A+B+ √(2)C+D)y^2+(A- √(2)B+C+ √(2)D)y+(B+D) )/(y^4+1)`


`\longrightarrow \left \{ {{A+C=0\to A=-C} \atop { -√(2)(A-C)+(B+D)=2 }} \right.`

`-√(2)(A-C)+(B+D)=2 \to - √(2) *2A+0=2\to A = - (1)/( √(2) )`

`A+C=0\to C=-A\to C = - (- (1)/( √(2) ) )\to C = (1)/( √(2) )`

`(A+C)+ √(2) (D-B)=0\to 0+ √(2) (D-B)=0 \to B=D=0`

`B+D=0`

5)pass: Adopt what was used above:


`I = \int\limits ( -(1)/( √(2) )y )/(y^2+ √(2)y+1 ) dy+ \int\limits ( (1)/( √(2) )y )/(y^2- √(2)y+1 ) dy`


`I = - (1)/( √(2) ) \underbrace{\int\limits (y)/(y^2+ √(2y)+1 )dy }_(I_1)+ (1)/( √(2) ) \underbrace{\int\limits (y)/(y^2- √(2y)+1 )dy }_(I_2)`

6) pass: Now, solve it separately


`I_1 = \int\limits (y)/(y^2+ √(2y)+1 )dy`

`2I_1 = \int\limits(2y)/(y^2+ √(2y)+1 )dy = \int\limits (2y- √(2)+ √(2) )/(y^2+ √(2y)+1 ) dy`

`2I_1 = \int\limits(2y+ √(2) )/(y^2+ √(2y)+1 )dy - \int\limits( √(2) )/(y^2+ √(2y)+1 )dy`

`2I_1 = ln|y^2+ √(2)y+1| - √(2) \int\limits (1)/((y √(2)+1)^2+1 )dy`

`\boxed - ( √(2) )/(2) arctan(y √(2)+1 )`

and


`I_2 = \int\limits (y)/(y^2- √(2y)+1 )dy`

`2I_2 = \int\limits(2y)/(y^2- √(2y)+1 )dy = \int\limits (2y- √(2)+ √(2) )/(y^2+ √(2y)+1 ) dy`

`2I_2 = \int\limits(2y- √(2) )/(y^2- √(2y)+1 )dy - \int\limits( √(2) )/(y^2- √(2y)+1 )dy`

`2I_2 = ln|y^2- √(2)y+1| + √(2) \int\limits (1)/((y √(2)-1)^2+1 )dy`

`\boxedI_2 = (1)/(2) ln`

7) pass: Let's use the expression of
`I`


`I = - (I_1)/( √(2) ) + (I_2)/( √(2) )`


`I = - ( (1)/(2)ln|y^2+ √(2)y+1|- ( √(2) )/(2)arctan(y √(2) + 1) )/( √(2) ) + ( (1)/(2)ln|y^2- √(2)y+1|+ ( √(2) )/(2)arctan(y √(2) - 1) )/( √(2) )`

`I = - (1)/(2 √(2) ) ln|y^2+ √(2) y +1|+(1)/( √(2) ) arctan(y √(2) +1)+`

`+(1)/(2 √(2) ) ln|y^2- √(2) y +1|+(1)/( √(2) ) arctan(y √(2)-1)`

8) pass: Now, returning to the expression as a function of x, we finally arrive at the final answer:


`I = - (1)/(2 √(2) ) ln|y^2+ √(2) y +1|+(1)/( √(2) ) arctan(y √(2) +1)+`

`+(1)/(2 √(2) ) ln|y^2- √(2) y +1|+(1)/( √(2) ) arctan(y √(2)-1)`

`I = - (1)/(2 √(2) ) ln|( √(tan\:x))^2+ √(2)( √(tan\:x)) +1| + (1)/( √(2) ) arctan(( √(tan\:x)) √(2) +1`

`+(1)/(2 √(2) ) ln|( √(tan\:x))^2- √(2)( √(tan\:x)) +1| + (1)/( √(2) ) arctan(( √(tan\:x)) √(2) -1`

Answer:


`\boxedI = - (1)/(2 √(2) ) ln`

`\boxed + (1)/( √(2) )arctan( √(2\:tan\:x) -1) +C}`
`\end{array}}\qquad\quad\checkmark`