## anonymous 5 years ago Evaluate these integrals using integration by parts? a) ∫ cos (In x) dx b) ∫ x tan^(2) x dx

1. atlchic

what u say first

2. anonymous

3. atlchic

tell me guess

4. anonymous

I'm not sure how to do these questions :(

5. watchmath

a) Let $$u=\cos(\ln x)$$ and $$dv=dx$$. So $$du=-\frac{1}{x}\sin(\ln x)dx$$ and $$v=x$$. So the integral is equal to $$uv-\int vdu$$ $$x\cos(\ln x)+\int\sin (\ln x)$$dx Do integration by parts one more time to get $$\int\cos(\ln x)\,dx=x\cos(\ln x)+x\sin(\ln x)-\int \cos(\ln x)\,dx$$ Then we have $$2\int\cos(\ln x)\,dx=x\cos(\ln x)+x\sin(\ln x)$$ Therefore $$\int\cos(\ln x)=\frac{x}{2}(\cos (\ln x)+\sin (\ln x))$$+C.

6. anonymous

second one i think is the same. $u=x$ $du=dx$ $dv = tan^2(x)$ $v=tan(x) -x$ $\int x tan^2x dx=x(tan(x)-x)-\int (tan(x)-x)dx$ $=xtan(x)-x^2+ln(cos(x))+\frac{x^2}{2}$