Integrate \[\tan x.{\sec ^2}xdx\].
Answer
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Hint: We use the concepts and techniques of trigonometry to solve this problem. Generally, integration is the reverse process of differentiation. We will also know about some substitution methods in integration to solve this problem. Knowing about the substitution methods helps us in solving many other problems in an easy way.
Complete step-by-step solution:
To integrate the given term, we need to use a substitution method.
In the question it is given that, we need to integrate \[\tan x.{\sec ^2}xdx\]
So, let \[I = \int {\tan x.{{\sec }^2}xdx} \]
Here, we all know that, \[\dfrac{d}{{dx}}\tan x = {\sec ^2}x\]
So, here, the derivative of \[\tan x\] also exists in the given question.
So, we need to substitute \[\tan x = t\] and on differentiating this on both sides, we get, \[{\sec ^2}xdx = dt\]
Now, substitute these values in the integration.
So, we get,
\[I = \int {t.({{\sec }^2}xdx) = \int {tdt} } \]
And we all know the formula, \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} \]
So, from this we get \[\int {tdt = \dfrac{{{t^{1 + 1}}}}{{1 + 1}}} = \dfrac{{{t^2}}}{2} + c\]
Now, substituting back the original value of \[t\] , we get,
\[I = \dfrac{{{{\tan }^2}x}}{2} + c\]
So, this is the required answer.
For verification, let's differentiate our answers and check whether we are getting the same question or not.
So, \[\dfrac{d}{{dx}}\left( {\dfrac{{{{\tan }^2}x}}{2} + c} \right) = \dfrac{d}{{dx}}\left( {\dfrac{{{{\tan }^2}x}}{2}} \right) + \dfrac{d}{{dx}}c\]
\[ = \dfrac{1}{2}\dfrac{d}{{dx}}{\tan ^2}x + 0\] -----as the derivative of a constant is 0.
\[= \dfrac{1}{2}(2)(\tan x)\dfrac{d}{{dx}}\tan x\] ----from the chain rule i.e., \[\dfrac{d}{{dx}}\left( {f(g(x))} \right) = f'(g(x)) \times g'(x)\]
\[= \tan x.{\sec ^2}x\]
So, we are getting a correct answer and finally, \[\int {\tan x.{{\sec }^2}xdx = \dfrac{{{{\tan }^2}x}}{2} + c} \]
Note: After integrating an indefinite integral, we should add an arbitrary constant for must. So, be careful that you add it. And if in an integral, there exists a term such that its derivative also exists in the integral, then take that term as a substitution, so that, after substitution, the derivative vanishes and the problem becomes easier.
Here, we took \[\tan x = t\] as a substitution because the derivative of \[\tan x\] exists in the integral which is \[{\sec ^2}x\].
Complete step-by-step solution:
To integrate the given term, we need to use a substitution method.
In the question it is given that, we need to integrate \[\tan x.{\sec ^2}xdx\]
So, let \[I = \int {\tan x.{{\sec }^2}xdx} \]
Here, we all know that, \[\dfrac{d}{{dx}}\tan x = {\sec ^2}x\]
So, here, the derivative of \[\tan x\] also exists in the given question.
So, we need to substitute \[\tan x = t\] and on differentiating this on both sides, we get, \[{\sec ^2}xdx = dt\]
Now, substitute these values in the integration.
So, we get,
\[I = \int {t.({{\sec }^2}xdx) = \int {tdt} } \]
And we all know the formula, \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} \]
So, from this we get \[\int {tdt = \dfrac{{{t^{1 + 1}}}}{{1 + 1}}} = \dfrac{{{t^2}}}{2} + c\]
Now, substituting back the original value of \[t\] , we get,
\[I = \dfrac{{{{\tan }^2}x}}{2} + c\]
So, this is the required answer.
For verification, let's differentiate our answers and check whether we are getting the same question or not.
So, \[\dfrac{d}{{dx}}\left( {\dfrac{{{{\tan }^2}x}}{2} + c} \right) = \dfrac{d}{{dx}}\left( {\dfrac{{{{\tan }^2}x}}{2}} \right) + \dfrac{d}{{dx}}c\]
\[ = \dfrac{1}{2}\dfrac{d}{{dx}}{\tan ^2}x + 0\] -----as the derivative of a constant is 0.
\[= \dfrac{1}{2}(2)(\tan x)\dfrac{d}{{dx}}\tan x\] ----from the chain rule i.e., \[\dfrac{d}{{dx}}\left( {f(g(x))} \right) = f'(g(x)) \times g'(x)\]
\[= \tan x.{\sec ^2}x\]
So, we are getting a correct answer and finally, \[\int {\tan x.{{\sec }^2}xdx = \dfrac{{{{\tan }^2}x}}{2} + c} \]
Note: After integrating an indefinite integral, we should add an arbitrary constant for must. So, be careful that you add it. And if in an integral, there exists a term such that its derivative also exists in the integral, then take that term as a substitution, so that, after substitution, the derivative vanishes and the problem becomes easier.
Here, we took \[\tan x = t\] as a substitution because the derivative of \[\tan x\] exists in the integral which is \[{\sec ^2}x\].
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