Question 1

Which of these integrals is equivalent to the given integral?

∫ x^3/(√9x^4+6x^2-1) dx

a. ∫ x^3/(3x^2+1) dx
b. 1/6 ∫ √ secθ-1 /√2tanθ dθ

Question 2

Answer:

$\mathbf{\int (x^3)/(√(9x^4+ 6x^2 -1)) dx \ \ is \ \ equivalent \ \ to \ \ (1)/(18) \int ( √(2)* sec^2 \theta - sec \theta) d\theta}$

Explanation:

$\int (x^3)/(√(9x^4+ 6x^2 -1)) dx = \int (x^3 \ dx)/(√((3x^22)^2 + 2(3x^2) (1) + 1-2))$

$\implies \int(x^3 \ dx)/(√((3x^2 +1)^2-2))$

let;

$3x^2 + 1 = √(2)\ sec \theta$

$6xdx = √(2) sec \theta tan \theta \ d \theta$

$xdx = (√(2))/(6) sec \theta tan \theta \ d \theta$

$\implies \int(x^2*x \ dx)/(√(2 \sec^2 \theta -2))$

$\implies \int((1)/(3)(√(2) \ sec \theta - 1)*(√(2))/(6) sec \theta tan \theta \ d\theta )/(√(2 ) \ tan \theta )$

$\implies (1)/(18) \int ( √(2)* sec \theta - 1) \ sec \theta d \theta$

$\implies (1)/(18) \int ( √(2)* sec^2 \theta - sec \theta) d\theta$

∴

$\mathbf{\int (x^3)/(√(9x^4+ 6x^2 -1)) dx \ \ is \ \ equivalent \ \ to \ \ (1)/(18) \int ( √(2)* sec^2 \theta - sec \theta) d\theta}$

Please log in or register to add a comment.

Please log in or register to answer this question.