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Question

Question: How do you integrate \(x{\sec ^2}x\) ?...

How do you integrate xsec2xx{\sec ^2}x ?

Explanation

Solution

For integrating xsec2xx{\sec ^2}x , this involve two functions, for this we have to use integration by parts that is , uvdx=uvdx(dudxvdx)dx\int {u \cdot vdx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} } } dx . Here the criteria for selecting first function uu is according to the ILATEILATE rule , where II denotes inverse Trigonometric Function, LL denotes Logarithm Function, AA denotes Algebraic Function, TT denotes Trigonometric Function,
EE denotes Exponential Function .

Formula used:
Integration by parts
uvdx=uvdx(dudxvdx)dx\int {u \cdot vdx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} } } dx
Here the criteria for selecting first function uu is according to the ILATEILATE rule , where II denotes inverse Trigonometric Function, LL denotes Logarithm Function, AA denotes Algebraic Function, TT denotes Trigonometric Function, EE denotes Exponential Function .

Complete step by step solution:
We have to find the integration of function xsec2xx{\sec ^2}x ,
Therefore, we can write as follow ,
I=xsec2xdxI = \int {x \cdot {{\sec }^2}xdx} (to find)
For integrating this we have to use integration by parts that is
uvdx=uvdx(dudxvdx)dx\int {u \cdot vdx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} } } dx
Here the criteria for selecting first function uu is according to the ILATEILATE rule , where II denotes inverse Trigonometric Function, LL denotes Logarithm Function, AA denotes Algebraic Function, TT denotes Trigonometric Function, EE denotes Exponential Function .
Here xx is an algebraic function and sec2x{\sec ^2}x is a trigonometric function. Algebraic function comes first.
Therefore, we take xx as our first function.
xsec2xdx=xsec2xdx(dxdxsec2x)dx\int {x \cdot {{\sec }^2}xdx = x\int {{{\sec }^2}xdx - \int {\left( {\dfrac{{dx}}{{dx}}\int {{{\sec }^2}x} } \right)dx} } }
We know that integration of sec2x{\sec ^2}x is equal to tanx\tan x ,
=xtanx1tanxdx= x\tan x - \int {1 \cdot \tan xdx}
We also know that integration of tanx\tan x is equal to logsecx\log \sec x ,
=xtanxlogsecx+c= x\tan x - \log \sec x + c
Hence , we have the required result.

Note: When doing indefinite integration, always write cc part after the integration. This cc 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.