Integrate 2-3sinx/cos^2x dx - Weebly
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Integrate 2-3sinx/cos^2x dx
1234567891011121314151617181920212223242526272829303132Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic ExpertsDedicated counsellor for each studentDetailed Performance Evaluationview all courses Last updated at Dec. 6, 2019 by Ex 7.1, 20 2 3 sin cos 2 2 3 sin cos 2 = 2 cos 2 3 sin cos 2 = 2 sec 2 3 sin cos . 1 cos = 2 sec 2 3 tan . sec =2 sec 2 3 tan sec = + Dear student I=2-3sinxcos2xdx=-3sinx-2cos2xdx=-3sinxcos2xdx+2cos2xdx=K+LConsider, K=-3sinxcos2xdxTake u=cosxdu=-sinxdxSo K=31u2du=-3u=-3cosxK=2cos2xdx=2sec2xdx=2 tanxSo I=-3cosx+2tanx+C Regards Here is an example of integration where you would use the chain rule. You would begin by integrating 3sin(x) to get -3cos(x) and then cos(2x) to get 1/2sin(2x). This is because whatever is multiplied by the x inside of the brackets affects what number goes in front of the sin(x), cos(x) etc. The reciprocal of that number is brought to the front and multiplied by the number that was already there. In this instance, the 2 in cos(2x) was made 1/2, and multiplied by the 1 at the front. The cos(2x) is then integrated to become 1/2sin(2x). I know that it's very easy to go wrong and make silly mistakes by mixing up the integration and differentiation of sin(x) and cos(x), so i personally always draw a diagram, if you will (drawn on whiteboard and explained). This is always the case whenever you have anything added together - you can integrate them separately and then put them all into the equation at the end - not forgetting the +c of course! So the answer to this question would be that 3sin(x) + cos(2x) integrates to 1/2sin(2x) - 3cos(x) +c.
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