নির্দিষ্ট যোগজ

A=acos2 θ \theta +bsin2 θ \theta  

  0π2dθA \int_{0}^{\frac{\pi}{2}} \frac{d \theta}{A} এর মান কোনটি?

অসীম স্যার

 0π2dθacos2θ+bsin2θ=0π2sec2θdθa+btan2θ \begin{array}{l} \text { } \int_{0}^{\frac{\pi}{2}} \frac{\mathrm{d} \theta}{\mathrm{acos}^{2} \theta+\mathrm{b} \sin ^{2} \theta} \\ =\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \theta \mathrm{d} \theta}{\mathrm{a}+\mathrm{btan}^{2} \theta} \end{array}

ধরি, tanθ=z \tan \theta=\mathrm{z}

sec2θdθ=dz এখন, 0dza+bz2=[1abtan1(baz)]0তখन θ=0 তখন z=0θ=π2 তখन z==1ab×π2=π2ab \begin{array}{l} \therefore \sec ^{2} \theta d \theta=d z \\ \text { এখন, } \int_{0}^{\infty} \frac{d z}{a+b z^{2}} \\ =\left.\left[\frac{1}{\sqrt{a b}} \tan ^{-1}\left(\sqrt{\frac{b}{a}} z\right)\right]_{0}^{\infty}\right|_{\text {তখन } \theta=0 \text { তখন } z=0} \theta=\frac{\pi}{2} \text { তখन } z=\infty \\ =\frac{1}{\sqrt{a b}} \times \frac{\pi}{2}=\frac{\pi}{2 \sqrt{a b}} \end{array}

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