Integrate (Sin(x)+Sin(x)^3)dx/Cos(2x).

Integrate (Sin(x)+Sin(x)^3)dx/Cos(2x).
You won't find this result by substitution. I used my Integrate to Product Rule.
Using "I" as the integration sign (what is the code for it ?), the result is:
I (sin x + sin^3 x)dx/cos 2x = ( ln (cos 2x) cos x )/4 +
(1/2) I (tan x) ln (cos 2x)dx/cos x  3/4 I (sin x) ln (cos2x) dx.
It doesn't look like you'd get away without an infinity of integrals.
This looks like homework. You can do it though, treat the Sin(x)/Cos(2x) and Sin(x)^3/Cos(2x) terms separately and consider the substitution u = Cos(x).Originally Posted by anandsatya
Also, you are going to need to use partial fractions so make sure you know how to do the required long division
This is my effort, for what it's worth.
Write expression as sinx(1 + sin<sup>2</sup>x)/cos2x.
Using sin<sup>2</sup>x + cos<sup>2</sup>x = 1
and cos<sup>2</sup>x  sin<sup>2</sup>x = cos2x,
rewrite the above expression as:
sinx(2  cos<sup>2</sup>x)/(2cos<sup>2</sup>x  1)
Change variable according to u = cosx, du = sinxdx.
Intergrand becomes:
(u<sup>2</sup>  2)du/(2u<sup>2</sup>  1).
By algebraic division,
(u<sup>2</sup>  2)/(2u<sup>2</sup>  1) = 1/2  3/2(2u<sup>2</sup> 1)
Integral of first term is u/2 ( or cosx/2)
Integral of second term is a standard integral:
(derived by partial fractions, as suggested by river_rat)
integral of du/(a<sup>2</sup>  u<sup>2</sup>) = (1/2a).ln{(a + u)/(a  u)}
I think !! ( Please check it out)
Looks fine by me old fool
Just don't forget the + C at the end...
Thank you Old Fool.
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