Question
Evaluate the following integrals:
$\int\frac{\cos\text{x}}{\sin^2\text{x}+4\sin\text{x}+5}\text{dx}$
$\int\frac{\cos\text{x}}{\sin^2\text{x}+4\sin\text{x}+5}\text{dx}$
✓
Answer
$\int\frac{\cos\text{x dx}}{\sin^2\text{x}+4\sin\text{x}+5}$
Let $\sin\text{x = t}$
$\Rightarrow\cos\text{x dx = dt}$
Now, $\int\frac{\cos\text{x dx}}{\sin^2\text{x}+4\sin\text{x}+5}$
$=\int\frac{\text{dt}}{\text{t}^2+4\text{t}+5}$
$=\int\frac{\text{dt}}{\text{t}^2+2\times\text{t}\times2+4+1}$
$=\int\frac{\text{dt}}{(\text{t}+2)^2+1^2}$
$=\frac{1}{1}\tan^{-1}\Big(\frac{\text{t}+2}{1}\Big)+\text{C}$
$=\tan^{-1}(\sin\text{x}+2)+\text{C}$
Let $\sin\text{x = t}$
$\Rightarrow\cos\text{x dx = dt}$
Now, $\int\frac{\cos\text{x dx}}{\sin^2\text{x}+4\sin\text{x}+5}$
$=\int\frac{\text{dt}}{\text{t}^2+4\text{t}+5}$
$=\int\frac{\text{dt}}{\text{t}^2+2\times\text{t}\times2+4+1}$
$=\int\frac{\text{dt}}{(\text{t}+2)^2+1^2}$
$=\frac{1}{1}\tan^{-1}\Big(\frac{\text{t}+2}{1}\Big)+\text{C}$
$=\tan^{-1}(\sin\text{x}+2)+\text{C}$
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