Ellen Chang

2022-07-02

Issues computing without using ${\mathrm{sin}}^{2}\left(x\right)=1-{\mathrm{cos}}^{2}\left(x\right)$

Elias Flores

Expert

It means you are going round and round on the answer . You can also use $\mathrm{cos}\left(2x\right)=2{\mathrm{cos}}^{2}\left(x\right)-1=\frac{1-{\mathrm{tan}}^{2}\left(x\right)}{1+{\mathrm{tan}}^{2}\left(x\right)}$ and then normal substitution of $\mathrm{tan}\left(x\right)=t$

ttyme411gl

Expert

Hint:
${\mathrm{cos}}^{2}x=\frac{1}{2}\mathrm{cos}\left(2x\right)+\frac{1}{2}$
$=\frac{1}{2}\int \mathrm{cos}\left(2x\right)dx+\frac{1}{2}\int 1dx$

Do you have a similar question?