Integrating this function

[matattack]

Given a function f(x)= \(\displaystyle (3x-\Pi)\cos\frac{1}{2}(x)\) and asked to asked to find the are enclosed by the graph and the x axis between -pi and pi

I am using integration by parts to integrate the function but Im not sure I have done it correctly the following are my workings


u = \(\displaystyle (3x-\pi)\)
V= \(\displaystyle \cos\frac{1}{2}x\)
using the integration by parts method
\(\displaystyle 3x-\pi 2\sin\frac{1}{2}x-\int 2\sin\frac{1}{2}x (3)\)
\(\displaystyle \rightarrow 6x-\pi \sin\frac{1}{2}x-6\int \sin\frac{1}{2}x\)
\(\displaystyle \rightarrow 6x-\Pi \sin\frac{1}{2}(-6x2)\cos\frac{1}{2}x \)
\(\displaystyle \rightarrow 6x-\pi\sin\frac{1}{2}-12\cos\frac{1}{2}x+c \)
I have tried this question a few times but have gotten different answers, has anyone idea where I have gone wrong ?

 

[Steven G]

Given a function f(x)= \(\displaystyle (3x-\Pi)\cos\frac{1}{2}(x)\) and asked to asked to find the are enclosed by the graph and the x axis between -pi and pi

I am using integration by parts to integrate the function but Im not sure I have done it correctly the following are my workings


u = \(\displaystyle (3x-\pi)\)
V= \(\displaystyle \cos\frac{1}{2}x\)
using the integration by parts method
\(\displaystyle 3x-\pi 2\sin\frac{1}{2}x-\int 2\sin\frac{1}{2}x (3)\)
\(\displaystyle \rightarrow 6x-\pi \sin\frac{1}{2}x-6\int \sin\frac{1}{2}x\)
\(\displaystyle \rightarrow 6x-\Pi \sin\frac{1}{2}(-6x2)\cos\frac{1}{2}x \)
\(\displaystyle \rightarrow 6x-\pi\sin\frac{1}{2}-12\cos\frac{1}{2}x+c \)
I have tried this question a few times but have gotten different answers, has anyone idea where I have gone wrong ?

Can you write down the formula you are trying to use and then try to use? BTW, ALL integral have a differential at the end of it (like dx, dt, dy,....)