Question

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

Answer 1

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.