1. ## Integral

How do I solve using substitution?

$\int\sin4xtanxdx$

2. $\int\sin4x\tan{x}~dx=\int\frac{\sin4x\sin{x}}{\cos {x}}~dx=\frac12\int\frac{\cos3x-\cos5x}{\cos{x}}~dx$

3. Originally Posted by Krizalid
$\int\sin4x\tan{x}~dx=\int\frac{\sin4x\sin{x}}{\cos {x}}~dx=\frac12\int\frac{\cos3x-\cos5x}{\cos{x}}~dx$
Thanks for the help. Can you explain the last step on how you got that...and...

I copied the problem incorrect it's:
$\int\sec4xtanxdx$

4. Originally Posted by xxpl510xx
Thanks for the help. Can you explain the last step on how you got that
Consider the following two identities:
$cos(a - b) = cos(a)cos(b) + sin(a)sin(b)$
$cos(a + b) = cos(a)cos(b) - sin(a)sin(b)$

Subtract the bottom from the top:
$cos(a - b) - cos(a + b) = 2sin(a)sin(b)$

Thus
$sin(a)sin(b) = \frac{1}{2}(cos(a - b) - cos(a + b))$

Thus
$sin(4x)sin(x) = \frac{1}{2}(cos(4x - x) - cos(4x + x))$

$sin(4x)sin(x) = \frac{1}{2}(cos(3x) - cos(5x))$

-Dan

5. Originally Posted by xxpl510xx
Thanks for the help. Can you explain the last step on how you got that...and...

I copied the problem incorrect it's:
$\int\sec4xtanxdx$
I love jokes. This one rocks!!

6. Originally Posted by xxpl510xx
Thanks for the help. Can you explain the last step on how you got that...and...

I copied the problem incorrect it's:
$\int\sec4xtanxdx$
I bet also that the new one is supposed to be
INT.[(secX)^4 tanX] dX

So,
= INT.[(secX)^3](secX tanX dX)