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as       Now the integral can be written as (B) is correct.
As we know  Using the above property we can write the integral as Answer (D) is correct
cos2x=cos2x-sin2x let sinx+cosx=t,(cosx-sinx)dx=dt hence the given integral can be written as B is correct
the above integral can be re arranged as let ex=t, exdx=dt   (A) is correct
As we know where b-a=hn In the given problem b=1, a=0 and
Integrating by parts we get For I2 take 1-x2 = t2, -xdx=tdt Hence Proved
The integral is written as Hence Proved
For I2 let cosx=t, -sinxdx=dt The limits change to 0 and 1 I1-I2=2/3 Hence proved.
The Integrand g(x) therefore is an odd function and therefore
L.H.S =  We can write the numerator as [(x+1) -x]       = RHS Hence proved.
Given integral  So, we split it in according to intervals they are positive or negative. Now,   as  is positive in the given x -range                       Therefore,    as  is in the given x -range  and  in the range  Therefore,    as  is in the given x -range  and  in the range  Therefore,  So, We have the sum
Let I =                -(i) Replacing x with ( -x),                    - (ii) Adding (i) and (ii)
Let I = Here, we can see that if we put sinx = t, then the whole function will convert in term of t with dx being changed to dt.so Now the important step here is to change the limit of the integration as we are changing the variable.so, So our function becomes, Now, let's integrate this by using integration by parts method,
First let's assume t = cosx - sin x so that (sinx +cosx)dx=dt So, Now since we are changing the variable, the new limit of the integration will be, when x = 0, t = cos0-sin0=1-0=1 when   Now, Hence our function in terms of t becomes,
First, let's get rid of the square roots from the denominator,
Here first let convert sin2x as the angle of x ( sinx, and cosx)  Now let's remove the square root form function by making a perfect square inside the square root Now let ,  since we are changing the variable, limit of integration will change our function in terms of t :
Lets first simplify the function.   As we have a good relation in between squares of the tan and square of sec lets try to take our equation there, AS we can write square of sec in term of tan,   Now let's calculate the integral of the second function, (we already have calculated the first function) let   here we are changing the variable so we have to calculate the limits of the new...
First, let's convert sin and cos into tan and sec. (because we have a good relation in tan and square of sec,) Let' divide both numerator and denominator by  Now lets change the variable the limits will also change since the variable is changing  So, the integration becomes:
Since, we have   multiplied by some function, let's try to make that function in any function and its derivative.Basically we want to use the property,   So, Here let's use the property so,
Here let's first reduce the log function. Now, let  So our function in terms if new variable t is : now let's solve this By using integration by parts
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