Hello

hope you are doing great. As per your query  I would glad to tell you that

this question can be solved by the formula log(1+x)/x when x tends to zero convert log argument(sinx) in the form of 1+x by adding and substractin 1 and then multiply numerator and denominator by sinx-1 . then whole log part will become 1 and you will be left with  sinx-1 on putting x=0 the result will be -1

Dear Candidate,

Due to our limited answering options & confusion while reading your question, it is difficult for us to give you an answer based on your statement. Kindly search the same query on various portals on internet & you will find a detailed step by step answer for your sum.

Thanks.

Integral (sin x) (cos x) dx

Let u = cos x

Then du = -sin x

Integral -u du = -u²/2 + C, where C is the constant of integration.

Integral -u du = -cos²x/2 + C

Putting limit from 0 to 90:

cos 90 = 0

cos 0 = 1

=> -cos²(90)/2+C - {-cos²(0)/2+C}

=> -0+C - {-1/2+C}

=> -0+C + 1/2-C

=> 1/2

You can check the list of colleges below

Government

Indian Institute of Technology Gandhinagar

Maharaja Sayajirao University of Baroda, Vadodara

Sardar Vallabhbhai National Institute of Technology Surat

Dharmsinh Desai University, Nadiad

LD College of Engineering, Ahmedabad

Private

Dhirubhai Ambani Institute of Information and Communication Technology, Gandhinagar

Nirma University, Ahmedabad

CEPT University, Ahmedabad

Charotar University of Science and Technology, Petlad

Pandit Deendayal Petroleum University, Gandhinagar

To check the complete list you can refer below link

https://engineering.careers360.com/colleges/ranking/2017/top-engineering-colleges-in-gujarat

Good Luck!!