Integration

∫cos⁡(ln⁡x) dx= ? \int \cos(\ln x) \, dx = \, ?

#Calculus

Note by Abdelfatah Teamah
4 years, 2 months ago

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Comments

Following @Pi Han Goh 's hint:

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

∫cos⁡(ln⁡x) dx=∫eucos⁡(u) du\displaystyle \int \cos( \ln x ) \,dx = \int e^u \cos(u) \,du

Integrating by parts twice gives

∫eucos⁡(u) du=cos⁡u∫eu du−∫(dducos⁡(u))(∫eu du) du=eucos⁡u+∫eusin⁡u du=eucos⁡u+sin⁡u∫eu du−∫(ddusin⁡(u))(∫eu du) du=eucos⁡u+eusin⁡u−∫eucos⁡u du\begin{aligned} \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{aligned}

∴2∫eucos⁡u du=eu(cos⁡u+sin⁡u)  ⟹  ∫eucos⁡u du=eu(cos⁡u+sin⁡u)2+C\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 uu by ln⁡x\ln x , we get

∫cos⁡(ln⁡x) dx=x2(cos⁡(ln⁡x)+sin⁡(ln⁡x))+C\displaystyle \int \cos ( \ln x ) \,dx = \dfrac{x}{2} \left( \cos (\ln x) + \sin (\ln x) \right) + C

Tapas Mazumdar - 4 years, 2 months ago

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

∫ex(f(x)+f′(x)) dx=exf(x)+C. \int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C .

Pi Han Goh - 4 years, 2 months ago

What have you tried? Where are you stuck on?

Hint: Let y=ln⁡x y = \ln x, then integration by parts.

Pi Han Goh - 4 years, 2 months ago
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