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Evaluate \( \int \frac{2 \cos x}{\left(\sin ^{2} x+1\right)\left(\sin ^{2} x+2\right)} d x \) |
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Answer» Let I = \(\int\frac{2cosx}{(sin^2x+1)(sin^2x+2)}dx\) Let sin x = t ⇒ cos x dx = dt \(\therefore\) I = \(\int\frac{2dt}{(t^2+1)(t^2+2)}\) \(=2\int\frac{(t^2+2)-(t^2+1)}{(t^2+1)(t^2+2)}dt\) \(=2\int(\frac1{t^2+1}-\frac1{t^2+2})dt\) \(=2[\int\frac1{t^2+1}dt-\int\frac1{t^2+2}dt)\) \(=2[tan^{-1}t-\frac1{\sqrt2}tan^{-1}\frac t{\sqrt2})+c\) (\(\because\) \(\int\frac{1}{a^2+x^2}dx=\frac1atan^{-1}\frac xa\)) \(=2tan^{-1}t-\sqrt2tan^{-1}\frac t{\sqrt2}+c\) \(=2tan^{-1}(sin x)-\sqrt2tan^{-1}(\frac{sin x}{\sqrt2})+c\) (\(\because\) t = sin x) |
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