Please help! I need to approximate the integral using a 3rd Order Taylor Series Expansion: `int_0^(pi/2) cos(x^2)dx` BUT! It must be done using the breaking up the integral and evaluating each...

Please help! I need to approximate the integral using a 3rd Order Taylor Series Expansion: `int_0^(pi/2) cos(x^2)dx`

BUT! It must be done using the breaking up the integral and evaluating each expression: 

`int_0^(pi/2) cos(x^2)dx = int_0^(pi/6) cos(x^2)dx + int_(pi/6)^(pi/3) cos(x^2)dx + int_(pi/3)^(pi/2) cos(x^2)dx`

I don't understand!! Can you also please help me understand better by explaining why this approach would be more accurate than evaluating the integral directly.

1 Answer | Add Yours

gsarora17's profile pic

gsarora17 | (Level 2) Associate Educator

Posted on

The taylor series expansion of

`cos(x)=sum_(n=0)^oo (-1)^nx^(2n)/(2n!)`

`cos(x^2)=sum_(n=0)^oo(-1)^n(x^2)^(2n)/(2n!)`

`cos(x^2)=1-x^4/(2!)+x^8/(4!)-x^12/(6!)+ ......`

Now considering expansion upto the third order,

`cos(x^2)=1-x^4/2+x^8/24`

Now we can integrate,

`int_0^(pi/2)cos(x^2)dx=int_0^(pi/2)(1-x^4/2+x^8/24)dx`

`int_0^(pi/2)cos(x^2)dx=[x-x^5/(5*2)+x^9/(9*24)]_0^(pi/2)`

Now evaluating the integral by plugging in the limits.

`int_0^(pi/2)cos(x^2)dx=[pi/2-pi^5/(2^5*5*2)+pi^9/(2^9*9*24)]`

Now plug in the value of pi,

`int_0^(pi/2)cos(x^2)dx=(22/14-22^5/(7^5*2^5*5*2)+22^9/(7^9*2^9*9*24))`

`int_0^(pi/2)cos(x^2)dx=(1.571428571-0.958237639+0.270519175)`

=0.883710107

We’ve answered 318,911 questions. We can answer yours, too.

Ask a question