int cos (log x) d x= | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $I = \int \cos (\log x) d x$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$. Then,$I = \int e^t \cos t dt$. Using the inte

Vedclass · Accepted Answer

$\frac{x}{2}(\cos (\log x)+\sin (\log x))+c$,(where $c$ is a constant of integration)

Vedclass · Answer

(C) Let $I = \int \cos (\log x) d x$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$. Then,$I = \int e^t \cos t dt$. Using the integration by parts formula $\int e^t f(t) dt = e^t \int f(t) dt - \int (e^t \int f(t) dt) dt$ or the standard result $\int e^t (\cos t + \sin t) dt = e^t \cos t + C$,we apply integration by parts: $I = \cos t \cdot e^t - \int (-\sin t) \cdot e^t dt = e^t \cos t + \int e^t \sin t dt$. Applying integration by parts again to $\int e^t \sin t dt$: $I = e^t \cos t + [\sin t \cdot e^t - \int \cos t \cdot e^t dt] = e^t \cos t + e^t \sin t - I$. Thus,$2I = e^t (\cos t + \sin t) + c_1$. $I = \frac{e^t}{2} (\cos t + \sin t) + c$. Substituting back $t = \log x$ and $e^t = x$: $I = \frac{x}{2} (\cos (\log x) + \sin (\log x)) + c$.

$\int \cos (\log x) d x=$

Similar Questions

Integrate the function: $\left(x^{2}+1\right) \log x$

Evaluate the integral: $\int \sin(\log x) dx$

$\int \tan^{-1} x \, dx = $

$\int (\log_{e} 2x)^3 dx =$

$\int \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right) d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module