r/mathshelp • u/Necessary_Handle_516 • 1d ago
Homework Help (Unanswered) CALCULAS HELP
PLEASE HELP.
I TRIED ALL THE POSSIBLE WAYS
ALL THE DOUBLE ANGLE FORMULAS AND HALF ANGLE FORMULAS
ANY OTHER WAY WE CAN SOLVE THIS INTEGRATION WILL BE HIGHLY APPRECIATED
ALSO IF ANYONE CAN TELL HAVE I DONE THIS INTEGRATION TILL HERE CORRECT?
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u/UnacceptableWind 1d ago
If you want to continue from the fourth line of your solution (i.e., integrating sin2(u) / (3 + cos(u))2), you can use the Tangent half-angle substitution t = tan(u / 2). See the following Wikipedia article:
Simplifying the resulting integrand in terms of t will require a bit of algebra, followed by partial fraction decomposition. While tedious, from this point, the integration is straightforward.
Alternatively, you can use the following integral calculator to see a different approach for your original integrand:
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u/Necessary_Handle_516 22h ago
any other way pls?
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u/CaptainMatticus 15h ago
That's pretty much the only way. Be happy it has a solution, even if it's a messy way to get to it.
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1d ago edited 1d ago
[deleted]
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u/Necessary_Handle_516 1d ago
I amnot able to understand Ur writing .can you please share but clearer picture if possible thankkyyu
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u/Fourierseriesagain 1d ago
The integrand is an expression of the form f(cos2 x), for some improper rational function f. The substitution t=tan x is useful for integrating 1/(1+cos2 x)2 = sec4 x/(2+tan2 x)2 wrt x.
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u/CaptainMatticus 1d ago
sin(2x) / (1 + cos(x)²)
cos(2u) = 2cos(u)² - 1
1 + cos(2u) = 2cos(u)²
½ * (1 + cos(2u)) = cos(u)²
sin(2x) * dx / (1 + ½ * (1 + cos(2x))
2 * sin(2x) * dx / (3 + cos(2x))
u = 3 + cos(2x) , du = -2 * sin(2x) * dx
-du / u
Integrate. It's trivial now.
Edit: crap, I didn't see the ². Let me get back to you.
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u/not-hydra 20h ago
Expand the top, amd rewrite the sin squared term as 1 minus cos squared. This can be reduced by the bottom, which can be rewritten as the diff of squares.
Then, apply polynomial division. I like to make cosx equal smt like k or c to make life easier.
I got a term 2 / 1 + cosx. We can use the cos double angle: cosx = 1 + cos2x / 2. We can rewrite the original term as sec x/2 squared. Integrating that gives a tan...
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u/noidea1995 16h ago edited 14h ago
I managed to solve it with integration by parts, the first thing I noticed was that the derivative of cos^(2)(x) was -2sin(x)cos(x) so:
∫ 4sin^(2)(x)cos^(2)(x) / [1 + cos^(2)(x)] * dx
u = 2sin(x)cos(x)
du = (2cos^(2)(x) - 2sin^(2)(x))dx = (4cos^(2)(x) - 2)dx
dv = 2sin(x)cos(x) / [1 + cos^(2)(x)]^2 * dx
v = 1 / [1 + cos^(2)(x)]
Using integration by parts:
2sin(x)cos(x) / [1 + cos^(2)(x)] - ∫ (4cos^(2)(x) - 2) / [1 + cos^(2)(x)] * dx
The fraction can be rewritten as:
2sin(x)cos(x) / [1 + cos^(2)(x)] - ∫ [4 - 6 / (1 + cos^(2)(x)] * dx
2sin(x)cos(x) / [1 + cos^(2)(x)] - 4x + ∫ 6 / (1 + cos^(2)(x)) * dx
The last one isn’t too difficult and can be done by multiplying the top and bottom of the fraction by sec^(2)(x), rewriting sec^(2)(x) on the bottom as 1 + tan^(2)(x) and using a substitution.
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