Integral big question

Question

Anyone could help me to solve this equation

I try to study lots of material but I coulden't solve it. I am a high school student and try to learn. $\displaystyle\int \cos(\ln(x))dx$?

Answer 1

Hint: try integration by parts. Note that $$ \int \cos(\ln x)\; dx= x\cos(\ln x)+\int \sin(\ln x)\; dx $$ And the same "trick" on the last integral will lead you to the answer.

Edit: As said before, integration by parts yields for the second integral $$ \int \sin(\ln x)\; dx= x\sin(\ln x)-\int \cos(\ln x)\; dx $$ So $$ \int \cos(\ln x)\; dx=x\sin(\ln x)+ x\cos(\ln x)-\int\cos(\ln x)\; dx $$ Therefore, $$ \int \cos(\ln x)\; dx=\frac{x}{2}(\sin(\ln x)+ \cos(\ln x)) $$

Answer 2

Let $u = \ln x$, then $x = e^u$, and $dx = e^udu$. Thus $I = \int e^u\cdot \cos udu$. This integral is popular in calculus textbook and you can find an answer.

Answer 3

Through integration by parts, $$\int \cos(\ln(x))dx $$

Let $u = \cos(\ln(x))$ and $dv = dx$, thus,

$$du = -\frac{\sin(ln(x))}{x}, v = x$$

Hence,

$$ \int \cos(\ln(x))dx = u \times v - \int du \times v\\ \int \cos(\ln(x))dx = \cos(\ln(x)) \times x + \int \sin(ln(x))dx $$

Since we have another awkward integral on the rhs, integrate it again, this time we will get the key to solution,

Let $ w = \sin(\ln(x))$ and $dz = dx$, thus $$dw = \cfrac{\cos(\ln(x))}{x} \ \text{and} \ z = x$$

Back to our solution,

$$ \int \cos(\ln(x))dx = \cos(\ln(x)) \times x + (\sin(\ln(x)) \times x - \int\cos(\ln(x))dx) $$

To get rid of the integral on the right hand side, we add the additive inverse of the rhs integral,

$$ 2\int \cos(\ln(x))dx = \cos(\ln(x)) \times x + \sin(\ln(x)) \times x\\ \int \cos(\ln(x))dx = \cfrac{\cos(\ln(x)) \times x + \sin(\ln(x) \times x)}{2}$$