int frac{sin x cdot cos x}{sin ^{4} x+cos ^{4 | vedclass

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(C) Let $I = \int \frac{\sin x \cdot \cos x}{\sin ^{4} x + \cos ^{4} x} dx$. Dividing the numerator and the denominator by $\cos ^{4} x$,we get: $I =

Vedclass · Accepted Answer

$\frac{1}{2} 	an ^{-1}(	an ^{2} x)+c$

Vedclass · Answer

(C) Let $I = \int \frac{\sin x \cdot \cos x}{\sin ^{4} x + \cos ^{4} x} dx$. Dividing the numerator and the denominator by $\cos ^{4} x$,we get: $I = \int \frac{\frac{\sin x \cdot \cos x}{\cos ^{4} x}}{\frac{\sin ^{4} x}{\cos ^{4} x} + 1} dx = \int \frac{	an x \sec ^{2} x}{	an ^{4} x + 1} dx$. Let $	an ^{2} x = t$. Then,differentiating both sides with respect to $x$,we get $2 	an x \sec ^{2} x dx = dt$,which implies $	an x \sec ^{2} x dx = \frac{1}{2} dt$. Substituting these into the integral: $I = \int \frac{1}{1 + t^{2}} \cdot \frac{1}{2} dt = \frac{1}{2} \int \frac{1}{1 + t^{2}} dt$. Using the standard integral formula $\int \frac{1}{1 + t^{2}} dt = 	an ^{-1} t + c$,we get: $I = \frac{1}{2} 	an ^{-1} t + c = \frac{1}{2} 	an ^{-1}(	an ^{2} x) + c$.

$\int \frac{\sin x \cdot \cos x}{\sin ^{4} x+\cos ^{4} x} d x=$

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