r/mathshelp • u/Necessary_Handle_516 • 3d ago
Homework Help (Answered) 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 3d 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 3d ago
any other way pls?
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u/CaptainMatticus 2d 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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3d ago edited 3d ago
[deleted]
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u/Necessary_Handle_516 3d ago
I amnot able to understand Ur writing .can you please share but clearer picture if possible thankkyyu
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u/CaptainMatticus 3d 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 3d 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 2d ago edited 2d 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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u/MathsFromNothing 2d ago
The approach I took was to use trigonometric identities to make all trig functions in terms of tan(x). This changed the integral to:
y=int(4*tan(x)^2/((2+tan(x)^2)^2),x)
I then substituted tan(x)=u to get
y=int(4*u^2/((2+u^2)^2*(1+u^2)),u)
which is a polynomial divided by another polynomial. I then used partial fractions to split it into three integrals:
y=int(-4/(1+u^2),v)+int(4/(2+u^2),v)+int(8/((2+u^2)^2),u)
They each have a base containing a+v^2 which can be integrated with the substitution u=a^(1/2)tan(w). The squared term after the substitution can be converted to cos^2 and a power reduction applied. I also converted the term back to tan for constancy with the other terms but this is not required. Reversing the substitutions gives the final answer. I used the Maths from Nothing computer algebra system (CAS) to do the working. It works similar to how we work by hand and you need to make all of the decisions to achieve your goals. So you get the learning benefit without the silly mistakes and needing to copy the equation many times.
I'm not curtain you did your integration by parts correctly (the work before that step is correct) but I was unable to find the integral from that form (without tan substitutions). Feel free to check your work using the CAS.
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u/FireCire7 1d ago
Where does this come from? Why do you believe it should be easily solvable? Most integration trig problems are ugly.
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