int frac{sin x}{sqrt{5 sin ^2 x+6 cos ^2 x}} | vedclass

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(D) Let $I = \int \frac{\sin x}{\sqrt{5 \sin ^2 x+6 \cos ^2 x}} \,d x$. Using $\sin ^2 x = 1 - \cos ^2 x$,we get: $I = \int \frac{\sin x}{\sqrt{5(1 -

Vedclass · Accepted Answer

$-\log \left(\cos x+\sqrt{\cos ^2 x+5}\right)+c$,where $c$ is the constant of integration

Vedclass · Answer

(D) Let $I = \int \frac{\sin x}{\sqrt{5 \sin ^2 x+6 \cos ^2 x}} \,d x$. Using $\sin ^2 x = 1 - \cos ^2 x$,we get: $I = \int \frac{\sin x}{\sqrt{5(1 - \cos ^2 x) + 6 \cos ^2 x}} \,d x$ $I = \int \frac{\sin x}{\sqrt{5 - 5 \cos ^2 x + 6 \cos ^2 x}} \,d x$ $I = \int \frac{\sin x}{\sqrt{5 + \cos ^2 x}} \,d x$. Let $u = \cos x$,then $du = -\sin x \,d x$,which implies $\sin x \,d x = -du$. Substituting these into the integral: $I = \int \frac{-du}{\sqrt{5 + u^2}} = -\int \frac{du}{\sqrt{(\sqrt{5})^2 + u^2}}$. Using the standard formula $\int \frac{dx}{\sqrt{a^2 + x^2}} = \log |x + \sqrt{a^2 + x^2}| + c$: $I = -\log |u + \sqrt{5 + u^2}| + c$. Substituting $u = \cos x$ back: $I = -\log |\cos x + \sqrt{5 + \cos ^2 x}| + c$.

$\int \frac{\sin x}{\sqrt{5 \sin ^2 x+6 \cos ^2 x}} \,d x=$

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