∫sin⁡x d(cos⁡x) is equal to - Infinity Learn by Sri Chaitanya

A
$\frac{1}{2} \sin 2 x - x + C$
B
$\frac{1}{2} (\frac{1}{2} \sin 2 x - x) + C$
C
$\frac{1}{2} (\frac{\sin 2 x}{2} + x) + C$
D
none of these

Solution:

We have,

$\begin{array}{l} d (\cos x) = \frac{d}{d x} (\cos x) \cdot d x = - \sin x d x \\ ∴ \int \sin x d (\cos x) = - \int \sin^{2} x d x = - \frac{1}{2} \int (1 - \cos 2 x) d x \\ \Rightarrow \int \sin x d (\cos x) = - \frac{1}{2} (x - \frac{\sin 2 x}{2}) + C \\ = \frac{1}{2} (\frac{1}{2} \sin 2 x - x) + C \end{array}$

$\int \sin x d (\cos x)$ is equal to

Solution:

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