Question

Integrate \[\int {x{{\sec }^2}} xdx\]

Answer 1

Hint: As we know that, \[Cosx\] is the ratio of base and hypotenuse and \[\operatorname{Sec} x\] is opposite to it. \[\cos x = \dfrac{{Base}}{{Hypotenuse}}\] Also, we can write, \[\sec x = \dfrac{1}{{\cos x}}\]
 Firstly we will solve questions by using integration by parts.
Apply the formula for the integration of u.v , then after substituting all the values we might use the following identities-
 \[{\sec ^2}x + {\operatorname{tan} ^2}x = 1\]
Following are the integration of functions-
 \[\int {{{\operatorname{sec} }^2}x} .dx = \tan x\]
Integration of \[\sec x = \left( { - \log \left| {\cos x} \right|} \right) + c\]
To solve this, we will use the above identity.

Complete step-by-step answer:
In this question we will use integration by parts.
 let \[I = \int {x.{{\sec }^2}} xdx\]
Now, we can apply the formula for integration two functions when they are in product, i.e. u.v , we get-
 \[\int {u.v = u.\int {v.dx - \int {\left( {\left( {\dfrac{{du}}{{dx}}} \right)\int {v.dx} } \right)} } } .dx\] ---- (i)
By using equation (i), we can solve the given integration.
Now, in the given question the value of u is x and v is \[{\sec ^2}x\] , So, we can solve this equation
We get-
 \[I\] = \[x\int {{{\sec }^2}xdx - \int {\left\{ {\dfrac{d}{{dx}}\left( x \right)\int {{{\sec }^2}xdx} } \right\}} } dx\]
Now, we know-
 \[\int {{{\operatorname{sec} }^2}x} .dx = \tan x\] and \[\dfrac{{dx}}{{dx}} = 1\]
After substituting all the values, we get-
 \[I\] = \[x\tan x - \left( { - \log \left| {\cos x} \right|} \right) + c\]
 \[I\] = \[x\tan x + \log \left| {\cos x} \right| + c\]
So, the correct answer is “ \[x\tan x + \log \left| {\cos x} \right| + c\] ”.

Note: While integrating we have to use the integration of u.v, while substituting the values, sometimes we get confused whether which terms we have to integrate or which terms we have to differentiate. So, use the below formula of integration of u.v very carefully. \[\int {u.v = u.\int {v.dx - \int {\left( {\left( {\dfrac{{du}}{{dx}}} \right)\int {v.dx} } \right)} } } .dx\] . The simple meaning of trigonometry is calculations of triangles. Also, in physics, trigonometry is used to find the components of vectors and also in projectile motion have a lot of application of trigonometry.