Integrasikan (1)/1+cotx menggunakan metode substitusi.

[ 1 / ( 1 + cos x/sin x ) ] dx

= [ ( sin x ) / ( sin x + cos x ) ] dx

= (1/2) • [ ( 2 sin x ) / ( sin x + cos x ) ] dx …

= (1/2) • { [ ( sin x + cos x ) – ( ​​cos x – sin x ) ] / ( sin x + cos x ) } dx …

= (1/2) • { 1 dx – [ ( cos x – sin x ) / ( sin x + cos x ) ] dx }

= (1/2) • { x – (1/u) du }, …

dimana … u = sin x + cos x

= ( x/2 ) – (1/2)· ln | kamu | + C

= ( x/2 ) – (1/2)· ln | sin x + cos x | + C

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