Question

integration of u. e^-u. sin4u​

Answer 1

Step-by-step explanation:

We have,

$let \: \: i = \int {e}^{ - u} . \sin(4u) du$

Using Integration By Parts, we have,

$i= {e}^{ - u} \int \sin(4u) du - \int( \frac{d}{du} ( {e}^{ - u} ) \int \sin(4u) du)du$

$i = {e}^{ - u} . \frac{ - \cos(4u) }{4} - \int - {e}^{ - u} . - \frac{ \cos(4u) }{4} du$

$i = - \frac{1}{4} .{e}^{ - u} . \cos(4u) - \frac{1}{4}. \int {e}^{ - u} \cos(4u) du$

$i = - \frac{1}{4} .{e}^{ - u} . \cos(4u) - \frac{1}{4}.( {e}^{ - u} \int \cos(4u) du - \int( \frac{d}{du} ( {e}^{ - u}) \int \cos(4u) du)du$

$i = - \frac{1}{4} .{e}^{ - u} . \cos(4u) - \frac{1}{8}. {e}^{ - u} \sin(4u) + \frac{1}{4} \int - {e}^{ - u} . \frac{ \sin(4u) }{4} du$

$i = - \frac{1}{4} .{e}^{ - u} . \cos(4u) - \frac{1}{8}. {e}^{ - u} \sin(4u) - \frac{1}{8} \int {e}^{ - u} \sin(4u) du$

$i = - \frac{1}{4} .{e}^{ - u} . \cos(4u) - \frac{1}{8}. {e}^{ - u} \sin(4u) + \frac{1}{8}i$

$i - \frac{i}{8} = - \frac{1}{4} .{e}^{ - u} . \cos(4u) - \frac{1}{8}. {e}^{ - u} \sin(4u)$

$\frac{7i}{8} = - \frac{1}{4} .{e}^{ - u} . \cos(4u) - \frac{1}{8}. {e}^{ - u} \sin(4u)$

$i = - \frac{2}{7} .{e}^{ - u} . \cos(4u) - \frac{1}{7}. {e}^{ - u} \sin(4u)$

$i = - \frac{1}{7}{e}^{ - u}.(2\cos(4u) + \sin(4u) )$