If f(x)=cos⁡x and g(x)=sin⁡x then ∫log⁡f(x)(f(x))2dx is

# If $f\left(x\right)=\mathrm{cos}x$ and $g\left(x\right)=\mathrm{sin}x$ then $\int \frac{\mathrm{log}f\left(x\right)}{\left(f\left(x\right){\right)}^{2}}\mathrm{d}x$ is

1. A

$f\left(x\right)\left(\mathrm{log}f\left(x\right)+1\right)+x+C$

2. B

$\frac{g\left(x\right)}{f\left(x\right)}\left(\mathrm{log}f\left(x\right)+1\right)+\frac{{x}^{2}}{2}+C$

3. C

$\frac{f\left(x\right)}{g\left(x\right)}\left(\mathrm{log}g\left(x\right)+1\right)+x+C$

4. D

$\frac{g\left(x\right)}{f\left(x\right)}\left[\mathrm{log}f\left(x\right)+1\right]-x+C$

Register to Get Free Mock Test and Study Material

+91

Live ClassesRecorded ClassesTest SeriesSelf Learning

Verify OTP Code (required)

### Solution:

$\begin{array}{l}I=\int \frac{\mathrm{log}\mathrm{cos}x}{{\mathrm{cos}}^{2}x}dx=\int {\mathrm{sec}}^{2}x\mathrm{log}\mathrm{cos}xdx\\ =\left(\mathrm{tan}x\right)\mathrm{log}\mathrm{cos}x-\int \mathrm{tan}x\left(\frac{-\mathrm{sin}x}{\mathrm{cos}x}\right)dx\\ =\left(\mathrm{tan}x\right)\mathrm{log}\mathrm{cos}x+\int \left({\mathrm{sec}}^{2}-1\right)dx\\ =\left(\mathrm{tan}x\right)\left[\mathrm{log}\mathrm{cos}x+1\right]-x+C\end{array}$

$=\frac{g\left(x\right)}{f\left(x\right)}\left[\mathrm{log}f\left(x\right)+1\right]-x+C$  