evalute limit x tends to zero log sin x?
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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
Number of integral values of x that satisfies x^xx^1/2=(xx^1/2)^x is
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.
Find integration of logs sin dx whose upper and lower limit are pi and zero
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.
an=integral pi/2to infinity e^-x.cosx^n DX then a4-a6/A4
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integral 0 to 90 min (sinx,cosx)dx
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