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# Integration

$\int \cos(\ln x) \, dx = \, ?$

Note by Abdelfatah Teamah
5 months ago

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Following @Pi Han Goh 's hint:

Let $$u = \ln x \implies du = \dfrac 1x dx$$, so our integral becomes

$\displaystyle \int \cos( \ln x ) \,dx = \int e^u \cos(u) \,du$

Integrating by parts twice gives

\begin{align} \int e^u \cos(u) \,du &= \cos u \int e^u \,du - {\large \int} \left( \dfrac{d}{du} \cos(u) \right) \left( \int e^u \,du \right) \,du \\ &= e^u \cos u + \int e^u \sin u \,du \\ &= e^u \cos u + \sin u \int e^u \,du - {\large \int} \left( \dfrac{d}{du} \sin(u) \right) \left( \int e^u \,du \right) \,du \\ &= e^u \cos u + e^u \sin u - \int e^u \cos u \,du \end{align}

$$\therefore \displaystyle 2 \int e^u \cos u \,du = e^u \left( \cos u + \sin u \right) \\ \implies \displaystyle \int e^u \cos u \,du = \dfrac{e^u \left( \cos u + \sin u \right)}{2} + C$$

Thus in our original integration, replacing $$u$$ by $$\ln x$$, we get

$\displaystyle \int \cos ( \ln x ) \,dx = \dfrac{x}{2} \left( \cos (\ln x) + \sin (\ln x) \right) + C$ · 4 months, 3 weeks ago

Great work! You can simplify your work if you apply one off the integration tricks, namely:

$\int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C .$ · 4 months, 3 weeks ago

What have you tried? Where are you stuck on?

Hint: Let $$y = \ln x$$, then integration by parts. · 5 months ago