integration of sinx show its proof
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∫ sin x dx let's use integration by parts
Let u = sin x ∴ du = cos x dx and dv = dx ∴ v = x
Substituting u and dv in the formula ∫ u dv = u•v - ∫ v du
∫ sin x dx = sin x x - ∫ x cos x dx
Again integrating ∫ x cos x dx using integration by parts, let u = x ∴ du= dx and dv = cos x dx ∴ v = sin x
= x sin x - [ x sin x - ∫ sin x dx ]
= x sin x - [ x sin x - ( - cos x ) ] + C
= x sin x -[ x sin x + cos x ]+ C.
= x sin x - x sin x - cos x + C. ( x sin x cancels out )
= - cos x + C
∴∫ sin x dx = - cos x + C
Let u = sin x ∴ du = cos x dx and dv = dx ∴ v = x
Substituting u and dv in the formula ∫ u dv = u•v - ∫ v du
∫ sin x dx = sin x x - ∫ x cos x dx
Again integrating ∫ x cos x dx using integration by parts, let u = x ∴ du= dx and dv = cos x dx ∴ v = sin x
= x sin x - [ x sin x - ∫ sin x dx ]
= x sin x - [ x sin x - ( - cos x ) ] + C
= x sin x -[ x sin x + cos x ]+ C.
= x sin x - x sin x - cos x + C. ( x sin x cancels out )
= - cos x + C
∴∫ sin x dx = - cos x + C
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