Question

Integrate the following function: $\frac{x^3 \sin(tan^{-1}x^4)}{1+ x^8}$

Answer 1

Answer:

Step-by-step explanation:

Concept:

I have applied change of variable method to find the integral of the given function.

$I=\int{\frac{x^3sin(tan^{-1}(x^4))}{1+x^8}}\:dx\\\\I=\int{sin(tan^{-1}(x^4))}.\frac{x^3}{1+x^8}}\:dx$

take,

$t=tan^{-1}(x^4)\\\\\frac{dt}{dx}=\frac{1}{1+(x^4)^2}.4x^3\\\\\frac{dt}{4}=\frac{1}{1+(x^4)^2}.x^3\:dx\\\\\frac{dt}{4}=\frac{1}{1+x^8}.x^3\:dx$

Now,

$I=\int{sint}\frac{dt}{4}\\\\I=\frac{1}{4}\int{sint}\:dt\\\\I=\frac{1}{4}cost+c\\\\I=\frac{1}{4}cos(tan^{-1}(x^4))+c$