int frac{cos 2 x cdot sin 4 x}{cos ^4 x(1+cos | vedclass

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(C) हमारे पास $I = \int \frac{\cos 2x \cdot 2 \sin 2x \cos 2x}{\cos^4 x (1 + \cos^2 2x)} dx$ है। चूंकि $\cos^2 x = \frac{1 + \cos 2x}{2}$,इसलिए $\cos^

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$\log \frac{(1+\cos 2 x)^2}{\left(1+\cos ^2 2 x\right)}+\sec ^2 x+c$

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(C) हमारे पास $I = \int \frac{\cos 2x \cdot 2 \sin 2x \cos 2x}{\cos^4 x (1 + \cos^2 2x)} dx$ है। चूंकि $\cos^2 x = \frac{1 + \cos 2x}{2}$,इसलिए $\cos^4 x = \frac{(1 + \cos 2x)^2}{4}$ है। इसे प्रतिस्थापित करने पर,$I = 4 \int \frac{\cos^2 2x \sin 2x}{(1 + \cos 2x)^2 (1 + \cos^2 2x)} dx$। मान लीजिए $t = \cos 2x$,तो $dt = -2 \sin 2x dx$,इसलिए $\sin 2x dx = -\frac{dt}{2}$। $I = 4 \int \frac{t^2}{(1+t)^2 (1+t^2)} \left(-\frac{dt}{2}\right) = -2 \int \frac{t^2}{(1+t)^2 (1+t^2)} dt$। आंशिक भिन्नों का उपयोग करते हुए,$\frac{t^2}{(1+t)^2 (1+t^2)} = \frac{1}{2(1+t)^2} - \frac{1}{2(1+t)} + \frac{t}{2(1+t^2)}$। $I = -2 \int \left[ \frac{1}{2(1+t)^2} - \frac{1}{2(1+t)} + \frac{t}{2(1+t^2)} \right] dt$। $I = -2 \left[ -\frac{1}{2(1+t)} - \frac{1}{2} \log |1+t| + \frac{1}{4} \log (1+t^2) \right] + c$। $I = \frac{1}{1+t} + \log |1+t| - \frac{1}{2} \log (1+t^2) + c$। चूंकि $1+t = 1+\cos 2x = 2\cos^2 x$,इसलिए $\frac{1}{1+t} = \frac{1}{2\cos^2 x} = \frac{1}{2} \sec^2 x$। अतः,$I = \sec^2 x + \log \frac{(1+\cos 2x)^2}{1+\cos^2 2x} + c$ प्राप्त होता है।

$\int \frac{\cos 2 x \cdot \sin 4 x}{\cos ^4 x(1+\cos ^2 2 x)} d x=$

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