Question

How do you evaluate the definite integral \[\int {\dfrac{{\cos x}}{{1 + \sin x}}} dx\] from \[\left[ {0,\dfrac{\pi }{2}} \right]\]

Answer 1

Hint: In order to solve this integral, first of all, we will assume \[1 + \sin x = t\] .After that we will differentiate it and transfer the given integral in terms of \[t\] .And then we will use the formula as, \[\int {\dfrac{{dx}}{x} = \ln } |x| + c\] to solve the given integral. And finally, we will substitute the new limits to get the required result.

Complete step-by-step answer:
In this question, we have to solve the given definite integral, \[\int {\dfrac{{\cos x}}{{1 + \sin x}}} dx\] from \[\left[ {0,\dfrac{\pi }{2}} \right]\]
which means we have to solve, \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\cos x}}{{1 + \sin x}}} dx\]
Let us consider the given integral as,
\[I = {\text{ }}\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\cos x}}{{1 + \sin x}}} dx{\text{ }} - - - \left( i \right)\]
Now, let us assume \[1 + \sin x = t{\text{ }} - - - \left( {ii} \right)\]
As we know that,
\[\dfrac{d}{{dx}}\left( c \right) = 0\]
\[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\]
So, by differentiating both the sides of equation \[\left( {ii} \right)\] we get
\[\left( {0 + \cos x} \right) = \dfrac{{dt}}{{dx}}\]
\[ \Rightarrow \left( {0 + \cos x} \right)dx = dt{\text{ }} - - - \left( {iii} \right)\]
So, by substituting the values from equation \[\left( {ii} \right)\] and equation \[\left( {iii} \right)\] in equation \[\left( i \right)\] we get,
\[I = {\text{ }}\int\limits_a^b {\dfrac{{dt}}{t}} {\text{ }}\]
Now, as we change the variable \[x\] from \[t\] ,we need to change the limits as well.
So, let us find out the value of the units \[a\] and \[b\] by substituting \[x = 0\] and \[x = \dfrac{\pi }{2}\] in equation \[\left( {ii} \right)\]
\[\therefore {\text{ }}a = 1 + \sin \left( 0 \right)\]
\[ \Rightarrow a = 1 + 0 = 1\]
And \[b = 1 + \sin \left( {\dfrac{\pi }{2}} \right)\]
\[ \Rightarrow b = 1 + 1 = 2\]
Therefore, we get
\[I = {\text{ }}\int\limits_1^2 {\dfrac{{dt}}{t}} {\text{ }}\]
Now, we know that
\[\int {\dfrac{{dx}}{x}} = \ln |x| + c\]
So, we get
\[I = {\text{ }}\left[ {\ln |t|} \right]_1^2\]
\[ \Rightarrow I = {\text{ }}\left[ {\ln |2| - \ln |1|} \right]\]
We know that \[\ln |1| = 0\]
So, \[I = {\text{ }}\left[ {\ln |2| - 0} \right]\]
\[ \Rightarrow I = \ln |2|\]
Hence, we get the value of the definite integral \[\int {\dfrac{{\cos x}}{{1 + \sin x}}} dx\] as \[\ln |2|\]
So, the correct answer is “\[\ln |2|\]”.

Note: In this question, many students solve the integral correctly but forget to change the limits. That is like in the above equation, they forget to transforms the limits \[0\] and \[\dfrac{\pi }{2}\] into \[1\] and \[2\] respectively and get the wrong answer even after doing the calculation correctly. So, this must be taken care of in the case of definite integrals.
Also, this question can be solved using a result, i.e., \[\int {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}dx = \ln |f\left( x \right)| + c} \]
As, the given integral is \[I = {\text{ }}\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\cos x}}{{1 + \sin x}}} dx{\text{ }} - - - \left( i \right)\]
Since, \[\dfrac{d}{{dx}}\left( {1 + \sin x} \right) = \cos x\]
Therefore, we get
\[I = {\text{ }}\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\cos x}}{{1 + \sin x}}} dx{\text{ }} = {\text{ }}\left[ {\ln |1 + \sin x|} \right]_0^{\dfrac{\pi }{2}}\]
\[ = \left[ {\ln \left( {1 + \sin \left( {\dfrac{\pi }{2}} \right)} \right)} \right] - \left[ {\ln \left( {1 + \sin \left( 0 \right)} \right)} \right]\]
We know that \[\ln |1| = 0\]
Therefore, we get
\[ = \left[ {\ln \left( {1 + 1} \right)} \right] - 0\]
\[ = \ln 2\] which is the required result.