Antiderivative issue

[dear2009]

Hello math help participants,


I am doing an antiderivative problem but I am having trouble finding how to do the derivative of this problem


14/(square root of 2)(cos^2)(7x)

Can anybody please show me the derivative of this part becuase I dont know how to start it.

Thanks in advance

 

[Deleted member 4993]

Hello math help participants,


I am doing an antiderivative problem but I am having trouble finding how to do the derivative of this problem


14/(square root of 2)(cos^2)(7x)

Can anybody please show me the derivative of this part becuase I dont know how to start it.

Thanks in advance

$\displaystyle \frac{d}{dx}\left (\frac{14}{\sqrt{2}}cos^2(x)\right )$

$\displaystyle = \frac{14}{\sqrt{2}}\cdot 2\cdot cos(x)\cdot \left (-sin(x)\right )$

Thats it ......

 

[BigGlenntheHeavy]

$\displaystyle Assuming \ this \ is \ what \ you \ want \ to \ integrate:$

$\displaystyle \int\frac{14}{\sqrt2}cos^{2}(7x)dx \ = \ 7\sqrt2\int cos^{2}(7x)dx$

$\displaystyle Let \ u \ = \ 7x, \ then \ du \ = \ 7dx, \ \frac{du}{7} \ = \ dx.$

$\displaystyle Ergo, \ we \ have \ \sqrt2 \int cos^{2}(u)du \ = \ \sqrt2 \int\frac{1+cos(2u)}{2}du$

$\displaystyle = \ \frac{\sqrt2}{2}\int[1+cos(2u)]du \ = \ \frac{\sqrt2}{2}\bigg[u+\frac{sin(2u)}{2}\bigg]+C$

$\displaystyle = \ \frac{\sqrt2}{2}[u+sin(u)cos(u)]+C, \ resubstituting \ we \ get: \ \frac{\sqrt2}{2}[7x+sin(7x)cos(7x)]+C.$

$\displaystyle Therefore, \ putting \ it \ all \ together, \ we \ get: \ \int\frac{14}{\sqrt2}cos^{2}(7x)dx \ = \ \frac{\sqrt2}{2}[7x+sin(7x)cos(7x)]+C$