If int frac{cos x}{sqrt{4 sin ^2 x+4 sin x+5} | vedclass

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Question

(B) Let $I = \int \frac{\cos x}{\sqrt{4 \sin ^2 x+4 \sin x+5}} d x$. Substitute $\sin x = t$, then $\cos x d x = d t$. Thus, $I = \int \frac{d t}{\sqr

Vedclass · Accepted Answer

$2 \sin x+1$

Vedclass · Answer

(B) Let $I = \int \frac{\cos x}{\sqrt{4 \sin ^2 x+4 \sin x+5}} d x$. Substitute $\sin x = t$, then $\cos x d x = d t$. Thus, $I = \int \frac{d t}{\sqrt{4 t^2+4 t+5}} = \frac{1}{2} \int \frac{d t}{\sqrt{t^2+t+\frac{5}{4}}}$. Completing the square in the denominator: $t^2+t+\frac{5}{4} = (t+\frac{1}{2})^2 + 1$. So, $I = \frac{1}{2} \int \frac{d t}{\sqrt{(t+\frac{1}{2})^2 + 1}}$. Using the formula $\int \frac{du}{\sqrt{u^2+a^2}} = \sinh^{-1}(\frac{u}{a}) + C$, we get: $I = \frac{1}{2} \sinh^{-1}(t+\frac{1}{2}) + C$. Substituting $t = \sin x$, we have $I = \frac{1}{2} \sinh^{-1}(\sin x + \frac{1}{2}) + C$. Comparing this with $\frac{1}{2} \sinh^{-1}(f(x)) + C$, we get $f(x) = \sin x + \frac{1}{2}$. Therefore, $2 f(x) = 2(\sin x + \frac{1}{2}) = 2 \sin x + 1$.

If $\int \frac{\cos x}{\sqrt{4 \sin ^2 x+4 \sin x+5}} d x=\frac{1}{2} \sinh ^{-1}(f(x))+C$, then find $2 f(x)$.

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