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Integrationof ?cotx

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ln C is Constant
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(integral) cot x dx = (integral) (cos x /sin x) dx set u = sin x. then we find du = cos x dx substitute du=cos x, u=sin x (integral) (cos x/sin x) dx = (integral) du/u solve integral = ln |u| + C substitute back u=sin x = ln |sin x| + C
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ln |sin x| + C
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let Y = (integral) cot x dx => (integral) (cos x /sin x) dx .........(1) put u = sin x. on differentiating u w.r.t x, we get, du = cos x dx substitute du=cos x dx and u=sin x in (1) (integral)(cos x/sin x)dx = (integral) du/u after integration, Y = log |u| + C substitute back...
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let Y = (integral) cot x dx => (integral) (cos x /sin x) dx .........(1) put u = sin x. on differentiating u w.r.t x, we get, du = cos x dx substitute du=cos x dx and u=sin x in (1) (integral)(cos x/sin x)dx = (integral) du/u after integration, Y = log |u| + C substitute back u=sin x Y = log |sin x| + C where C is a constant is the answer. read less
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Strategy: Make in terms of sin's and cos's; Use Subtitution. (integral) cot x dx = (integral) cos x sin x dx set u = sin x. then we find du = cos x dx substitute du=cos x, u=sin x (integral) cos x sin x dx = (integral) du u solve integral = ln |u| + C substitute back...
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Strategy: Make in terms of sin's and cos's; Use Subtitution. (integral) cot x dx = (integral) cos x sin x dx set u = sin x. then we find du = cos x dx substitute du=cos x, u=sin x (integral) cos x sin x dx = (integral) du u solve integral = ln |u| + C substitute back u=sin x = ln |sin x| + C therefore, (integral) cot x = ln|sin x| + C. read less
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