Math Help - Undefinite Integral Question

1. Undefinite Integral Question

I want to know if :

$\int [cos(x)\cdot e^{2sin(x))}+\pi \cdot cos(4x)+\sqrt{2}\cdot x^{1/3}+\frac{1}{x}] dx$

is equal to :

${\int cos(x)dx} \cdot {\int e^{2sin(x)}dx} + {\int \pi \cdot cos(4x)dx+{\int \sqrt{2}}}\cdot {x^{1/3}dx}+{\int \frac {1}{x}}dx$

and if it is not , how can i solve this indefinite integral

2. Originally Posted by Larrioto
I want to know if :

$\int [cos(x)\cdot e^{2sin(x))}+\pi \cdot cos(4x)+\sqrt{2}\cdot x^{1/3}+\frac{1}{x}] dx$

is equal to :

${\int cos(x)dx} \cdot {\int e^{2sin(x)}dx} + {\int \pi \cdot cos(4x)dx+{\int \sqrt{2}}}\cdot {x^{1/3}dx}+{\int \frac {1}{x}}dx$

and if it is not , how can i solve this indefinite integral
No, not quite. It's equal to:

${\int cos(x)e^{2sin(x)}\,dx} + {\int \pi \cdot cos(4x)dx+{\int \sqrt{2}}}\cdot {x^{1/3}dx}+{\int \frac {1}{x}}dx$

You can do the first term by substitution. Notice that cos(x) is the derivative of sin(x).

$u = sin(x)$

$\frac{du}{dx} = cos(x)$

$du = cos(x)dx$

${\int e^{2u}\,du} + \pi {\int cos(4x)dx+\sqrt{2}{\int }}\cdot {x^{1/3}dx}+{\int x^{-1}}dx$

3. would the answer be :

$\frac {e^{2cos(x)}}{2}+\pi \cdot(\frac {-sin{4x}}{4})+\sqrt{2}\cdot \frac{(3x^{\frac{4}{3}})}{4}+ln|x|$

4. Originally Posted by Larrioto
would the answer be :

$\frac {e^{2cos(x)}}{2}+\pi \cdot(\frac {-sin{4x}}{4})+\sqrt{2}\cdot \frac{(3x^{\frac{4}{3}})}{4}+ln|x|$
i think the final answer would be this:

$\frac {e^{2sen(x)}}{2}+\pi \cdot(\frac {sin{4x}}{4})+\sqrt{2}\cdot \frac{(3x^{\frac{4}{3}})}{4}+ln|x|
$

5. Originally Posted by PanZerOlas
i think the final answer would be this:

$\frac {e^{2sen(x)}}{2}+\pi \cdot(\frac {sin{4x}}{4})+\sqrt{2}\cdot \frac{(3x^{\frac{4}{3}})}{4}+ln|x|
$
Thank you