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प्रश्न
\[\int\frac{\cos x}{\sin^2 x + 4 \sin x + 5} dx\]
बेरीज
उत्तर
` ∫ { cos x dx}/{sin^2 x + 4\sin x + 5}`
\[\text{ let }\sin x = t\]
\[ \Rightarrow \text{cos x dx }= dt\]
Now, ` ∫ { cos x dx}/{sin^2 x + 4\sin x + 5}`
\[ = \int\frac{dt}{t^2 + 4t + 5}\]
\[ = \int\frac{dt}{t^2 + 2 \times t \times 2 + 4 + 1}\]
\[ = \int\frac{dt}{\left( t + 2 \right)^2 + 1^2}\]
\[ = \frac{1}{1} \tan^{- 1} \left( \frac{t + 2}{1} \right) + C\]
\[ = \tan^{- 1} \left( \sin x + 2 \right) + C\]
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