Integral Calculus Question 263
Question: $ \int_{{}}^{{}}{\frac{\tan x}{\sec x+\tan x}\ dx=} $
Options:
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 $
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Answer:
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}{ \sec x+\tan x }=\sec x\tan x+{{\sec }^{2}}x $ $ =\sec x(\sec x+\tan x)=\frac{\sec x}{\sec x-\tan x} $
$ \Rightarrow \frac{d}{dx}{ \sec x-\tan x+x+c }=\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} $ .
BETA
