Rajasthan Boardहिन्दी माध्यमकक्षा 12 साइन्सगणितसमाकलन2 Marks
Question
समाकलन ज्ञात कीजिए: $\int \frac{\sin x}{(1+\cos x)^{2}} d x$
✓
Answer
माना $I = \int \frac{\sin x}{(1+\cos x)^{2}} d x$
माना $1 + \cos x = t \Rightarrow -\sin x =\frac{d t}{d x}$
$\Rightarrow d x =\frac{d t}{-\sin x}$
$\therefore I=\int \frac{\sin x}{(1+\cos x)^{2}} d x =\int \frac{\sin x}{t^{2}} \times \frac{d t}{-\sin x} =-\int \frac{1}{t^{2}} d t$
$= -\int t^{−2}dt =\frac{-t^{-2+1}}{-2+1}+C =\frac{1}{t}+C=\frac{1}{1+\cos x} + C$
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