Evaluate $\int\cos(\ln x^2)dx$

Question

$$\int\cos(\ln x^2)dx$$

I've learned substitution method, integration by parts, and some other basic methods for integration, but I have no idea how to start this question.

How should I start this question?

Answer 1

For ease, let's write $\ln x^2 = 2 \ln x$. If you do the substitution $u = 2 \ln x$, so that $\frac{du}{dx} = \frac{2}{x}$ and $x = e^{u/2}$, then you end up with

$$\frac 12 \int e^{u/2} \cos u \;\mathrm{d}u,$$

which can be done quickly with $2$ integration by parts (and which is a classical integration by parts problem).

Answer 2

Using the fact that $\cos \theta=\text{Re }e^{i\theta}$,

$$ \begin{align*} \int \cos(\log x^2)\; dx &= \text{Re}\int e^{2i \log x} \; dx \\ &= \text{Re} \int x^{2i}\; dx \\ &= \text{Re}\left(\frac{x^{1+2i}}{1+2i} \right) \\ &= \frac{x}{5}\text{Re} \left( x^{2i} (1-2i)\right) \\ &= \frac{x}{5}\text{Re} \left( (\cos(2\log x)+i\sin(2\log x) ) (1-2i)\right)\\ &= x\frac{ \cos(2\log x)+2 \sin(2\log x)}{5}+C \end{align*} $$

Answer 3

taking $t=\ln x$ you get $dx=e^t dt$ $$\int\cos(\ln x^2)dx=\int\cos(2t)e^t dt=:I$$ Integrating by parts twice you get $$I=\int\cos(2t)e^t dt=e^t\cos(2t)+2\int\sin(2t)e^t dt+C=$$ $$=e^t\cos(2t)+2\left(e^t\sin(2t) - 2\int\cos(2t)e^t dt\right)+C=$$ $$=e^t\cos(2t)+2e^t\sin(2t)-4I +C\ .$$ Thus, $$I=\frac{1}{5}\left(e^t\cos(2t)+2e^t\sin(2t)\right)+C=\frac{1}{5}\left(x\cos(\ln x^2)+2x\sin(\ln x^2)\right)+C$$

Answer 4

Rewrite $$ \int\cos(\ln x^2)dx=\int\cos(2\ln x)dx. $$ Using IBP by taking $u=\cos(2\ln x)\;\Rightarrow\;du=-\dfrac2x\sin(2\ln x)\ dx$ and $dv=dx\;\Rightarrow\;v=x\,$ yields $$ \int\cos(2\ln x)dx=x\cos(2\ln x)+2\int\sin(2\ln x)dx+C_1.\tag1 $$ Again using IBP by taking $u=\sin(2\ln x)\;\Rightarrow\;du=\dfrac2x\cos(2\ln x)\ dx$ and $dv=dx\;\Rightarrow\;v=x$ yields $$ \int\sin(\ln x^2)dx=x\sin(2\ln x)-2\int\cos(2\ln x)dx+C_2.\tag2 $$ Plug in $(2)$ to $(1)$ yields \begin{align} \int\cos(2\ln x)dx&=x\cos(2\ln x)+2x\sin(2\ln x)-4\int\cos(2\ln x)dx+K\\ 5\int\cos(2\ln x)dx&=x\cos(\ln x^2)+2x\sin(\ln x^2)+K. \end{align}