int frac{sin x cos x}{sqrt{1-sin ^{4} x}} d x | vedclass

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Question

(A) Let $I = \int \frac{\sin x \cos x}{\sqrt{1-\sin ^{4} x}} dx$. Substitute $\sin^{2} x = t$. Differentiating both sides with respect to $x$,we get $

Vedclass · Accepted Answer

$\frac{1}{2} \sin ^{-1}(\sin ^{2} x)+C$

Vedclass · Answer

(A) Let $I = \int \frac{\sin x \cos x}{\sqrt{1-\sin ^{4} x}} dx$. Substitute $\sin^{2} x = t$. Differentiating both sides with respect to $x$,we get $2 \sin x \cos x dx = dt$,which implies $\sin x \cos x dx = \frac{1}{2} dt$. Substituting these into the integral,we get $I = \int \frac{1}{2 \sqrt{1-t^{2}}} dt$. Using the standard integral formula $\int \frac{1}{\sqrt{1-t^{2}}} dt = \sin^{-1} t + C$,we obtain $I = \frac{1}{2} \sin^{-1} t + C$. Finally,substituting $t = \sin^{2} x$ back,we get $I = \frac{1}{2} \sin^{-1}(\sin^{2} x) + C$.

$\int \frac{\sin x \cos x}{\sqrt{1-\sin ^{4} x}} d x$ is equal to

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