Question

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

Answer 1

what u say first

Answer 2

Please? :)

Answer 3

tell me guess

Answer 4

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

Answer 5

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.

Answer 6

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