Evaluate ∫02π1(1+acos⁡θ)2dθ, 0≤a&lt;1

Question

Evaluate  ∫02π1(1+acos⁡θ)2dθ, 0≤a&amp;lt;1

Louis Page · Accepted Answer

It is not difficult to check that for any b&amp;gt;1 we have:J(b)=∫02πdθb+cos⁡θ=4∫0π2dϕb+cos⁡(2ϕ)=4∫0+∞dt(1+t2)(b−1+2cos2(arctan⁡t))=2πb2−1hence it follows that:−J′(b)=∫02πdθ(b+cos⁡θ)2=2bπ(b2−1)32and by taking a=1b we get:∀a∈(0,1), ∫02πdθ(1+acos⁡θ)2=2(1−a2)32

Cheryl King · Accepted Answer

We can actually evaluate this integral indefinitely.This is the most basic thing I could come up with,Considerf(x)=sin⁡x1+acos⁡x⇒f′(x)=1+cos⁡x(1+acos⁡x)2Now integrate both sides,⇒f(x)=∫(cos⁡x+a)dx(1+acos⁡x)2Now integrate both sides,⇒f(x)=∫(cos⁡x+a)dx(1+acos⁡x)2⇒f(x)=1a∫(acos⁡x+1)dx(1+acos⁡x)2+a2−1a∫dx(1+acos⁡x)2⇒sin⁡x1+acos⁡x=1a∫dx1+acos⁡x+a2−1a⋅INow, &#039;I&#039; is the integral we wanted to evaluate and the only problem left is∫dx1+acos⁡xwhich is easily calculated by putting cos⁡x=1−tan2x21+tan2x2∫dx1+acos⁡x=21−a1−a1+aarctan⁡(t1−a1+a)+CWhere t=tan⁡x2

star233 · Accepted Answer

∫02πdθ[1+acos⁡(θ)]2=∫−ππdθ[1−acos⁡(θ)]2=−2ddb∫0πdθb−acos⁡(θ)|b=1=−2ddb[∫0π/2dθb−acos⁡(θ)+∫0π/2dθb+acos⁡(θ)]b=1=−4ddb[b∫0π/2dθb2−a2cos2⁡(θ)]b=1=−4ddb[b∫0π/2sec2⁡(θ)dθb2tan2⁡(θ)+b2−a2]b=1=−4ddb(b∫0∞dtb2t2+b2−a2)b=1=−4ddb(1b2−a2∫0∞dtt2+1)b=1=−2πddb(1b2−a2)|b=1=2πb(b2−a2)3/2|b=1=2π(1−a2)3/2

Evaluate \int_0^{2\pi}\frac{1}{(1+a\cos\theta)^2}d\theta,\ 0\leq a<1

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New Questions in Calculus 1