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The function f(θ)=ddθ∫0θdx1−cosθcosx satisfies the differential equation
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[a] we have f(θ)=ddθ∫0θdx1−cosθcosx =11−cos2θ=cosec2θ (using Leibnitz?s Rule) ⇒df(θ)dθ=−2cosec2θcotθ ⇒df(θ)dθ+2f(θ)cotθ=0
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| Question Text | The function f(θ)=ddθ∫0θdx1−cosθcosx satisfies the differential equation |
| Answer Type | Text solution:1 |
| Upvotes | 150 |
