Question

Evaluate \[\int {\dfrac{{\cos \,2x + \,2\,{{\sin }^2}x}}{{{{\cos }^2}x}}\,dx} \]

Answer 1

Hint: According to the given question, firstly break the value of \[(\cos \,2x)\] into its constituent equal form that is \[(1 - \,2\,{\sin ^2}x)\] . After simplifying the equation, try to use substitution method that is \[u = \tan \,x\] and differentiate in terms of x with the help of division rule that is \[\left[ {\dfrac{{B\dfrac{{dA}}{{dx}}\, - \,A\dfrac{{dB}}{{dx}}}}{{{B^2}}}} \right]\] and substitute the value of dx in terms of du . Then solve the integration to get the desired result.

Complete step-by-step solution:
As, it is given
\[\int {\dfrac{{\cos \,2x + \,2\,{{\sin }^2}x}}{{{{\cos }^2}x}}\,dx} \]
Performing individual operations,
\[\int {\dfrac{{\cos \,2x + \,2\,{{\sin }^2}x}}{{{{\cos }^2}x}}\,dx} \]
Substituting the formula, \[\left[ {\cos \,2x = 1 - \,2\,{{\sin }^2}x} \right]\]
We get,
\[ = \,\,\,\int {\,\dfrac{{1 - \,2\,{{\sin }^2}x\, + \,2\,{{\sin }^2}x\,}}{{{{\cos }^2}x}}\,dx} \]
After cancelling the similar terms in numerator we get,
\[ = \,\int {\dfrac{1}{{{{\cos }^2}}}x\,dx} \] eq. (1)
Let, \[u = \tan \,x\] eq. (2)
And as we know that \[\tan \,x = \dfrac{{\sin \,x}}{{\cos \,x}}\]
Therefore, \[u = \dfrac{{\sin \,x}}{{\cos \,x}}\]
Differentiating \[u\] with respect to \[x\] and applying the rule of differentiation which is
\[\left[ {\dfrac{{B\dfrac{{dA}}{{dx}}\, - \,A\dfrac{{dB}}{{dx}}}}{{{B^2}}}} \right]\] also known as division rule.
Here, as it clear \[A = \sin x\] and \[B = \cos x\]
So, \[\dfrac{{du}}{{dx}} = \,\dfrac{{\cos \,x.(\cos \,x)\, - \,\sin \,x( - \sin \,x)}}{{{{\cos }^2}x}}\]
After simplifying the equation we get,
\[ = \,\dfrac{{{{\cos }^2}x\, + \,{{\sin }^2}x}}{{{{\cos }^2}x}}\]
After applying the identity \[\left[ {\,{{\cos }^2}x\, + \,{{\sin }^2}x = 1} \right]\] we get,
\[\dfrac{{du}}{{dx}} = \dfrac{1}{{{{\cos }^2}x}}\]
On taking dx on right hand side we get,
\[\therefore \,\,du = \dfrac{1}{{{{\cos }^2}x}}\,dx\] eq. (3)
Putting the value of \[du\] in place of \[\dfrac{1}{{{{\cos }^2}x}}\] that is putting the value of eq. (2) in eq. (3)
\[ = \,\int {\dfrac{1}{{{{\cos }^2}x}}\,dx} \] [Initially]
\[ = \int {du} \]
As we know, \[\int {dx} \, = x\]
So we get, \[u + c\,\]
On replacing u in terms of x we get,
\[ = \tan \,x + c\] [From eq. (2)]

Therefore, the value of \[\int {\dfrac{{\cos \,2x\, + \,2\,{{\sin }^2}x}}{{{{\cos }^2}x}}\,dx} \] is \[(\tan \,x + \,c)\], where c is the constant whose value is not defined and which we get only after indefinite integration.

Note: To solve these types of questions, as we do not have any \[\dfrac{u}{v}\] rule or any \[u.v\] rule in integration to solve it directly. So, we use a substitution method to solve it in a simplified way as shown above in the solution. Don’t forget to write the c after indefinite integration because without c, the solution will be considered as the definite integration.