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The value of $\int \frac{\sin x+\cos x}{3+\sin 2 x} d x$, is |
$\frac{1}{4} \log \left(\frac{2+\sin x-\cos x}{2-\sin x+\cos x}\right)+C$ $\frac{1}{2} \log \left(\frac{2+\sin x}{2-\sin x}\right)+C$ $\frac{1}{4} \log \left(\frac{1+\sin x}{1-\sin x}\right)+C$ none of these |
$\frac{1}{4} \log \left(\frac{2+\sin x-\cos x}{2-\sin x+\cos x}\right)+C$ |
We have, $I =\int \frac{\sin x+\cos x}{3+\sin 2 x} d x$ $\Rightarrow I =\int \frac{1}{4-(1-\sin 2 x)} d(\sin x-\cos x)$ $\Rightarrow I =\int \frac{1}{2^2-(\sin x-\cos x)^2} d(\sin x-\cos x)$ $\Rightarrow I=\frac{1}{4} \log \left(\frac{2+\sin x-\cos x}{2-\sin x+\cos x}\right)+C$ |
