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Question Number 20386 by tammi last updated on 26/Aug/17
∫1−a2x2dx
Answered by $@ty@m last updated on 26/Aug/17
Letx=sinθa⇒dx=1acosθdθ⇒I=1a∫1−sin2θcosθdθ=1a∫cos2θdθ=12a∫(1+cos2θ)dθ=12a[θ+sin2θ2]+C=12a[sin−1ax+ax1−a2x2]+C
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