Very good.

@gayansamarasekara

28 күн бұрын

Of course. Thanks for the comment…!

@7:15 aren't you multiplying top & bottom by zero? So the later integrals need justification

@gayansamarasekara

Ай бұрын

Thanks for your comment. You indeed have a point since the integral is evaluated between (pi/4) and (3pi/4), where (pi/2) is in between, and sin(pi/2)=1. The integral may be split into two pieces between (pi/4) to (pi/2) and (pi/2) to (3pi/4) prior to the multiplication, and then evaluated in this case. The final answer still stays the same. However, the discontinuity at (pi/2) is a removable discontinuity. Therefore it does not affect the integral. Ideally, it should've been mentioned in the discussion itself. I am glad you brought it up here. Thanks.

I would have used the substitution x = u + pi/2, so the interval of integration would then be (-pi/2, pi/2)

@gayansamarasekara

Ай бұрын

Nice...! In fact the new limits will be (-pi/4, pi/4) in that case...!

You, sir, are a master😁

@gayansamarasekara

Ай бұрын

Thank you for your kind encouragement. 😊👍🏻.

The real question becomes how often is a teacher going to give students those specified bounds of integration that allow the student to use the identity sin(pi-x)= sin(x) and Sin(2pi-x)=sin(-x)=-sin(x). This theorem only helps when the bounds allow you to do a phase shift. In general integration by parts works with or without bounds so it’s better.

@gayansamarasekara

Ай бұрын

Thanks for your comment. I agree with you to a great extent. It depends on the teacher’s thought process in day to day gen ed calculus classes. However, for a good selection of challenging exams in some parts of the world, such questions are often given. Those exam markers make sure there are no easy alternatives available other than the direct application of the theorem. (For e.g. the question given at the end of the video is significantly easier that way rather than by parts, though the question discussed in the video can be easily solved using by parts as well - still it takes longer to solve that way). The reason is, just using one question, a student’s ability to simplify, knowledge on integration and trigonometry are all examined. Many applied sciences have special functions with special limits, and those are worth evaluating. For example, the normal distribution (statistics), can be integrated between specific limits, though there is no closed form. Also, there are many other functions, which are not trig functions, that can be created to be solved by direct application of the theorem. But overall, you have a practically valid point about most of the routine calculus exams given in schools.

@moeberry8226

Ай бұрын

@@gayansamarasekara I also agree with you to a large extent. Keep up the amazing videos brother. When I first looked at the integrand I thought of Feynman’s Trick. But then seen the + 1 added to sin(x) in the denominator and said it can be done without it.

@gayansamarasekara

Ай бұрын

@@moeberry8226 Thank you so much. Feynman...! Love it, I'll make one for that too. Thanks for your comment. Have a wonderful weekend.

@moeberry8226

Ай бұрын

@@gayansamarasekara you too brother. Inshallah you have a good weekend.