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