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

X has only one Integral value which satisfies the relation.

X=1 satisfies the relation. It should be Noted X cannot be negative since it under Root.

Another Value of x which satisfies the equation is X=9/4, however this is not an integral value.

0 cannot be considered as a Solution as 0^0 is undefined and not 1.

Hello,

We have to find the integration of logs. sinx.dx,

Intg.( logs.sinx.dx) = clearly logs is a constant, so take it out of the integration.

= logs. intg.( sinxdx) = - logs. cosx , now we have to put the upper and lower limits as pi and 0.

So we get,

- logs. ( -1 - 1 )

= 2 .logs

Hence this is the answer of the integration.

Hope it helps you.

Dear Candidate,

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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