Have been given this question and am strugling abit.
Any help appreciated:
Find the work done (W) by a mechanism between the angular displacement (x) of 0 and pie/2, by intergrating the following equation:
2cos(2x) + 3sin(3x)
Thanks
Have been given this question and am strugling abit.
Any help appreciated:
Find the work done (W) by a mechanism between the angular displacement (x) of 0 and pie/2, by intergrating the following equation:
2cos(2x) + 3sin(3x)
Thanks
Hello
$\displaystyle W=\int_{0}^{\frac{\pi}{2}}2\cos(2x)+3\sin(3x)\,\ma thrm{d}x=\int_{0}^{\frac{\pi}{2}}2\cos(2x)\,\mathr m{d}x+\int_{0}^{\frac{\pi}{2}}3\sin(3x)\,\mathrm{d }x$
As $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}(2x)=2$, one can writes $\displaystyle 2\cos(2x)=u'(x)\cdot \cos(u(x))$ for $\displaystyle u(x)=2x$. Can you find an anti-derivative of this function ? (think about the chain rule : $\displaystyle (v\circ u)'=u'\cdot v'\circ u$. You know what is $\displaystyle u$ and just need to find $\displaystyle v$)
The same idea applies for the other integral.
$\displaystyle 2\cos (2x)+3\sin (3x)$
Integration, cos goes to sin and sin goes to -cos.
Also, remember the chain rule for inside the brackets, differentiate the bracket and divide by what you get.
$\displaystyle \int{2\cos (2x)}=\frac{2\sin (2x)}{2}=\sin (2x)$ - (cancel the 2s)
$\displaystyle \int{3\sin (3x)=-\frac{3\cos (3x)}{3}}=-\cos (3x)$ - (cancel the 3s)
Answer: $\displaystyle \int_{0}^{\frac{\pi }{2}}{2\cos (2x)+3\sin (3x)=\left[ \sin (2x)-\cos (3x) \right]_{0}^{\frac{\pi }{2}}}$