A) \[\sec x+\tan x-x+c\]
B) \[\sec x-\tan x+x+c\]
C) \[\sec x+\tan x+x+c\]
D) \[-\sec x-\tan x+x+c\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\frac{\tan x}{(\sec x+\tan x)}\,dx}=\int_{{}}^{{}}{\frac{\tan x(\sec x-\tan x)}{(\sec x+\tan x)(\sec x-\tan x)}\,dx}\] Multiplying \[{{N}^{r}}\] and \[{D}'\] by \[(\sec x-\tan x),\] we get \[=\int_{{}}^{{}}{\frac{\tan x(\sec x-\tan x)}{({{\sec }^{2}}x-{{\tan }^{2}}x)}\,dx}=\int_{{}}^{{}}{(\sec x\tan x-{{\tan }^{2}}x)\,dx}\] \[=\int_{{}}^{{}}{\sec x\tan x\,dx}-\int_{{}}^{{}}{({{\sec }^{2}}x-1)\,dx}\] \[=\int_{{}}^{{}}{\sec x\tan x\,dx}-\int_{{}}^{{}}{{{\sec }^{2}}x\,dx}+\int_{{}}^{{}}{1\,dx}\] \[=\sec x-\tan x+x+c\]. Trick : By inspection, \[\frac{d}{dx}\left\{ \sec x+\tan x \right\}=\sec x\tan x+{{\sec }^{2}}x\] \[=\sec x(\sec x+\tan x)=\frac{\sec x}{\sec x-\tan x}\] \[\Rightarrow \frac{d}{dx}\left\{ \sec x-\tan x+x+c \right\}=\sec x\tan x-{{\sec }^{2}}x+1\] \[=-\sec x(\sec x-\tan x)+1=\frac{-\sec x}{\sec x+\tan x}+1=\frac{\tan x}{\sec x+\tan x}\].
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