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The value of dx1+ecosx must be same as: -
(a) 11e2tan1(1e1+etanx2)+c, (e lies between 0 and 1)
(b) 21e2tan1(1e1+etanx2)+c, (e lies between 0 and 1)
(c) 1e21loge21sinx1+ecosx+c, (e is greater than 1)
(d) 2e21loge+cosx+e21sinx1+ecosx+c, (e is greater than 1)

Answer
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Hint: Use the conversion formula: - cosx=(1tan2x21+tan2x2) and send (1+tan2x2) to the numerator. In the numerator use the identity, (1+tan2x2)=sec2x2. Now, assume tanx2=k in the denominator. Differentiate both the sides to find dx in terms of dk. Convert the integral in the form dxa2+x2, whose solution is 1atan1(xa). Finally, substitute the value of k to get the correct option.

Complete step by step answer:
We have been given: -
I=dx1+ecosx
Using the conversion: - cosx=(1tan2x21+tan2x2), we get,
I=dx1+e(1tan2x21+tan2x2)I=(1+tan2x2)dx(1+tan2x2)+e(1tan2x2)I=(1+tan2x2)dx(1+e)+(1e)tan2x2
Using the identity: - (1+tan2x2)=sec2x2, we get,
I=sec2x2dx(1+e)+(1e)tan2x2
Substituting, tanx2=k, we have,
d(tanx2)=dk12sec2x2dx=dksec2x2dx=2dk
Substituting these values in (i), we get,
I=2dk(1+e)+(1e).k2
I=2(1e)dk(1+e1e)+k2
This can be written as: -
I=2(1e)dk(1+e1e)2+k2
The above integral is of the form: - dxa2+x2 whose solution is 1atan1(xa).
I=21e×11+e1etan1(k1+e1e)+c, c = constant of integration.
I=21e×1e1+etan1(1e1+ek)+c
Substituting the value of k and simplifying, we get,
I=21e×1+e×tan1(1e1+etanx2)+cI=21e2×tan1(1e1+etanx2=k)+c
Now, for 1+e1e to be defined, e must be less than 1 and greater than 0. This is because the term inside the square root must not be negative.

So, the correct answer is “Option B”.

Note: One may note that we do not have to convert (1+tan2x2)=sec2x2 in denominator because we have to assume tanx2=k in denominator. If we will use this conversion in the denominator then we will not be able to solve the question. Finally, remember that we have to define the range of ‘e’ so that the term 1e1+e can be defined.
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