int frac{2 cos 2 x}{(1+sin 2 x)(1+cos 2 x)} d | vedclass

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Question

(D) ધારો કે $I = \int \frac{2 \cos 2 x}{(1+\sin 2 x)(1+\cos 2 x)} d x$.નિત્યસમનો ઉપયોગ કરતા: $\cos 2x = \frac{1-	an^2 x}{1+	an^2 x}$,$\sin 2x = \fra

Vedclass · Accepted Answer

$2 \log (1+	an x)-	an x+c$

Vedclass · Answer

(D) ધારો કે $I = \int \frac{2 \cos 2 x}{(1+\sin 2 x)(1+\cos 2 x)} d x$.નિત્યસમનો ઉપયોગ કરતા: $\cos 2x = \frac{1-	an^2 x}{1+	an^2 x}$,$\sin 2x = \frac{2	an x}{1+	an^2 x}$,અને $1+\cos 2x = \frac{2}{1+	an^2 x}$.$I = \int \frac{2 \left(\frac{1-	an^2 x}{1+	an^2 x}ight)}{\left(1+\frac{2	an x}{1+	an^2 x}ight) \left(\frac{2}{1+	an^2 x}ight)} d x$$I = \int \frac{1-	an^2 x}{(1+	an x)^2} d x = \int \frac{(1-	an x)(1+	an x)}{(1+	an x)^2} d x$$I = \int \frac{1-	an x}{1+	an x} d x$.અહીં $	an x = t$ લેતા,$\sec^2 x d x = d t$ થાય,પરંતુ આ પદાવલિ માટે સીધું સાદુંરૂપ આપતા:$I = \int \frac{1-t}{1+t} d t = \int (-1 + \frac{2}{1+t}) d t$$I = -t + 2 \ln(1+t) + c = 2 \ln(1+	an x) - 	an x + c$.

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

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