Question

How do you evaluate \[\int\dfrac{\text{sin}x}{1 + \cos^{2}x}dx\] from \[\left\lbrack \dfrac{\pi}{2},\ \pi \right\rbrack\] ?

Answer 1

Hint:In this question, we need to evaluate the given expression \[\int\dfrac{\text{sin}x}{1 + \cos^{2}x}dx\] also given the limits of the expression \[\left\lbrack \dfrac{\pi}{2},\ \pi \right\rbrack\] . Let us consider the given expression as \[I\].Then by using a method of substitution for integration we can substitute \[\cos x\] as \[t\]. We also need to note the limits of the integration , the limits of \[x\] are given as \[\dfrac{\pi}{2}\] to \[\pi\] if we substitute \[t\] in \[I\] , then the new limit for \[t\] also varies. Then we need to find the bounds of \[t\] . Finally we can apply the new limits and find the value of the given integrand.

Complete step by step answer:
Given, \[\int\dfrac{\text{sin}x}{1 + \cos^{2}x}dx\]
Here we need to evaluate the given expression from \[\left\lbrack \dfrac{\pi}{2},\ \pi \right\rbrack\]. Let us consider the given expression as \[I\] .
\[\Rightarrow \ I = \int_{\dfrac{\pi}{2}}^{\pi}{\dfrac{\text{sin}x}{1 + \cos^{2}x}dx }\] ••• (1)
Where the upper limit of integration is \[\pi\] and the lower limit is \[\dfrac{\pi}{2}\] .
Let us consider \[t = cos\ x\]
Now on differentiating \[t\] with respect to \[x\] ,
\[\Rightarrow \dfrac{dt}{dx} = - \sin\ x\]
On rearranging the terms,
We get,
\[\Rightarrow \ - dt = \sin\ x\ dx\]

Given that \[x\] varies from \[\dfrac{\pi}{2}\] to \[\pi\] .
Now we need to find the bounds of \[t\] .
By substituting \[x = \dfrac{\pi}{2}\] in \[t = \cos\ x\]
\[\Rightarrow \ t = \cos\left( \dfrac{\pi}{2} \right)\]
We know that the value of \[\cos\left( \dfrac{\pi}{2} \right)\] is \[0\].
Thus we get \[t = 0\]
Then we need to substitute \[x = \pi\] in \[t = \cos\ x\]
\[\Rightarrow \ t = \cos(\pi)\]
We know that the value of \[\cos(\pi)\] is \[- 1\]
Thus the limits of the integration vary from \[0\] to \[- 1\] .
Thus by rewriting the terms and limits, equation (1) becomes,
\[\Rightarrow \ I = \int_{0}^{- 1}{- \dfrac{1}{1 + t^{2}} dt}\]

Now on integrating,
We get,
\[I = - \left\lbrack \tan^{- 1}\left( t \right) \right\rbrack_{0}^{- 1}\]
Now on applying the limits inside,
We get,
\[\Rightarrow \ I = - \left( \tan^{- 1}\left( - 1 \right) - \tan^{- 1}\left( 0 \right) \right)\]
We know that \[\tan^{- 1}\left( - 1 \right)\] is \[- \dfrac{\pi}{4}\] and \[\tan^{- 1}\left( 0 \right)\] is \[0\] .
\[\Rightarrow \ I = - \left( - \dfrac{\pi}{4} – 0 \right)\]
On simplifying,
We get,
\[\therefore \ I = \dfrac{\pi}{4}\]

Thus the value of \[\int\dfrac{\text{sin}x}{1 + \cos^{2}x}dx\] from \[\left\lbrack \dfrac{\pi}{2},\ \pi \right\rbrack\] is \[\dfrac{\pi}{4}\].

Note:We must have a strong grip over the integral calculus to solve such a complex question of definite integration. Also while solving the definite integrals, it is important to make sure that we have to change the values of the limits of the integral otherwise, we may get the value of the integral wrong. We need to know that the indefinite integral gives us a family of curves whereas the definite integral gives a numeric value. We must be careful while doing the calculations in order to get the final answer.