The integral \int \, cos(loge x) dx is equal to : (where C is a constant of integration)

x \over 2 [cos(loge x) + sin(loge x)] + C

x \over 2 [sin(loge x) - cos(loge x)] + C

x[cos(loge x) + sin(loge x)] + C

x[cos(loge x) - sin(loge x)] + C

Indefinite Integrals PYQ — JEE Main 2019 · 12 Jan · Shift 1 · Q38 with Solution

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Correct answer: C

So the integral becomes $I=\int e^t\cos t\,dt.$

Now use the standard result $\int e^{at}\cos bt\,dt=\frac{e^{at}}{a^2+b^2}(a\cos bt+b\sin bt)+C.$

Here, $a=1$ and $b=1$ , so

=\frac{e^t}{2}(\cos t+\sin t)+C.$$ 4. Substitute back $e^t=x$ and $t=\ln x$: $$I=\frac{x}{2}\left[\cos(\ln x)+\sin(\ln x)\right]+C.$$ 5. Hence the correct option is $$\boxed{\text{C}}.$$ 6. Verification by differentiation: Let $$F(x)=\frac{x}{2}\left[\cos(\ln x)+\sin(\ln x)\right].$$ Then $$F'(x)=\frac{1}{2}\left[\cos(\ln x)+\sin(\ln x)\right]+ \frac{x}{2}\left[-\sin(\ln x)\cdot \frac{1}{x}+\cos(\ln x)\cdot \frac{1}{x}\right].$$ So $$F'(x)=\frac{1}{2}\left[\cos(\ln x)+\sin(\ln x)-\sin(\ln x)+\cos(\ln x)\right] =\cos(\ln x).$$ This matches the integrand exactly. 7. Comparison with stored answer: Stored correct answer is C, and our derived answer is also C. Therefore, they agree.