Question

The indefinite integral can be found in more than one way. First use the substitution method to find the indefinite integral. Then find it without using substitution. Check that your answers are equivalent. 6x^5(x^6-2)dx

Answer 1

Answer:

∫$6x^5(x^6-2)\,dx$ = $\frac{1}{2}(x^6-2)^2+C$

Step-by-step explanation:

To find:

∫$6x^5(x^6-2)\,dx$

Solution:

Method of substitution:

Let $x^6-2=t$

Differentiate both sides with respect to $t$

$6x^5\,dx=dt$

[use $(x^n)'=nx^{n-1}$]

So,

∫$6x^5(x^6-2)\,dx$ = ∫ $t\,dt$ = $\frac{t^2}{2}+C_1$ where $C_1$ is a variable.

(Use ∫$t^n\,dt=\frac{t^{n+1} }{n+1}$ )

Put $t=x^6-2$

∫$6x^5(x^6-2)\,dx$ = $\frac{1}{2}(x^6-2)^2+C_1$

Use $(a-b)^2=a^2+b^2-2ab$

So,

∫$6x^5(x^6-2)\,dx$ = $\frac{1}{2}(x^6-2)^2+C_1=\frac{1}{2}(x^{12}+4-4x^6)+C_1=\frac{x^{12} }{2}-2x^6+2+C_1=\frac{x^{12} }{2}-2x^6+C$

where $C=2+C_1$

Without using substitution:

∫$6x^5(x^6-2)\,dx$ = ∫$6x^{11}-12x^5\,dx$ = $\frac{6x^{12} }{12}-\frac{12x^6}{6}+C=\frac{x^{12} }{2}-2x^6+C$

So, same answer is obtained in both the cases.