Question
Question: How do you integrate \(x{\sec ^2}x\) ?...
How do you integrate xsec2x ?
Solution
For integrating xsec2x , this involve two functions, for this we have to use integration by parts that is , ∫u⋅vdx=u∫vdx−∫(dxdu∫vdx)dx . Here the criteria for selecting first function u is according to the ILATE rule , where I denotes inverse Trigonometric Function, L denotes Logarithm Function, A denotes Algebraic Function, T denotes Trigonometric Function,
E denotes Exponential Function .
Formula used:
Integration by parts
∫u⋅vdx=u∫vdx−∫(dxdu∫vdx)dx
Here the criteria for selecting first function u is according to the ILATE rule , where I denotes inverse Trigonometric Function, L denotes Logarithm Function, A denotes Algebraic Function, T denotes Trigonometric Function, E denotes Exponential Function .
Complete step by step solution:
We have to find the integration of function xsec2x ,
Therefore, we can write as follow ,
I=∫x⋅sec2xdx (to find)
For integrating this we have to use integration by parts that is
∫u⋅vdx=u∫vdx−∫(dxdu∫vdx)dx
Here the criteria for selecting first function u is according to the ILATE rule , where I denotes inverse Trigonometric Function, L denotes Logarithm Function, A denotes Algebraic Function, T denotes Trigonometric Function, E denotes Exponential Function .
Here x is an algebraic function and sec2x is a trigonometric function. Algebraic function comes first.
Therefore, we take x as our first function.
∫x⋅sec2xdx=x∫sec2xdx−∫(dxdx∫sec2x)dx
We know that integration of sec2x is equal to tanx ,
=xtanx−∫1⋅tanxdx
We also know that integration of tanx is equal to logsecx ,
=xtanx−logsecx+c
Hence , we have the required result.
Note: When doing indefinite integration, always write c part after the integration. This c part indicates the constant part remains after integration and can be understood when you explore it graphically. Infinite integration constant gets cancelled out, so we only write it in indefinite integration.