1. ## integrals

I can not figure out how to do this question. Do I seperate it into its sides first or something? Do I treat the nuber in front with the x as a constant? Im confused. Please show me what to do.

integrate (7x^3 cos(2+x^4) - 8x^3 e^(2+x^4))dx

2. ## Re: integrals

Hello, Twinkie!

I cannot figure out how to do this question.
Do I seperate it into its sides first or something? . What does that mean?
Do I treat the number in front with the x as a constant? . Of course!
I'm confused. . So am I.
Please show me what to do.

$\displaystyle \int \left[7x^3\cos(x^4+2) - 8x^3e^{x^4+2}\right]dx$

We have two integrals: .$\displaystyle \begin{Bmatrix}7\int x^3\cos(x^4+2)\,dx & [1] \\ \\ -8\int x^3e^{x^4+2}\,dx & [2] \end{Bmatrix}$

$\displaystyle [1]\;\text{Let }u \,=\,x^4+2 \quad\Rightarrow\quad du \,=\, 4x^3\,dx \quad\Rightarrow\quad x^3\,dx \,=\,\tfrac{1}{4}du$

Substitute: .$\displaystyle 7\int\cos(x^4+2)\left(x^3\,dx\right) \;=\;7\int\cos u\left(\tfrac{1}{4}\,du\right)$
. . . . . . . . $\displaystyle =\;\tfrac{7}{4}\int \cos u\,du \;=\;\tfrac{7}{4}\sin u + C_1$

Back-substitute: .$\displaystyle \tfrac{7}{4}\sin(x^4+2) + C_1$

$\displaystyle [2]\;\text{Let }u \,=\,x^4+2 \quad\Rightarrow\quad du \,=\, 4x^3\,dx \quad\Rightarrow\quad x^3\,dx \,=\,\tfrac{1}{4}du$

Substitute: .$\displaystyle -8\int e^{x^4+2}\left(x^3\,dx\right) \;=\;-8\int e^u\left(\tfrac{1}{4}\,du\right)$
. . . . . . . . $\displaystyle =\;-2\int e^u\,du \;=\;-2e^u + C_2$

Back-substitute: .$\displaystyle -2e^{x^4+2} + C_2$

Answer: .$\displaystyle \tfrac{7}{4}\sin(x^4+2) - 2e^{x^4+2} + C$

3. ## Re: integrals

What part of the equaision did the X^3 dx=(1/4)du come from?

4. ## Re: integrals

Originally Posted by Twinkie
What part of the equaision did the X^3 dx=(1/4)du come from?

Here let:

$\displaystyle u = x^4 + 2$

If you take the derivative of the above assumption with respect to $\displaystyle x$ you'll get:

$\displaystyle \frac{\mathrm{d}u}{\mathrm{d}x} = 4x^3$

Then multiply both sides of this by $\displaystyle \tfrac{1}{4}\,\mathrm{d}x$ and you'll get:

$\displaystyle \frac{1}{4}\,\,\mathrm{d}u = x^3\,\,\mathrm{d}x$

And hence:

$\displaystyle x^3\,\,\mathrm{d}x = \frac{1}{4}\,\,\mathrm{d}u$