Match List I with List II
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LIST I |
LIST II |
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A. $\int\frac{\sin x}{1+\cos x}dx$ |
I. $e^{\tan^{-1}x}+C$ |
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B. $\int\frac{1}{1-\tan x}dx$ |
II. $\log(\log x+1)+C$ |
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C. $\int\frac{e^{\tan^{-1}x}}{1+x^2}dx$ |
III. $-\log|1+\cos x|+C$ |
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D. $\int\frac{1}{x+x\log x}dx$ |
IV. $\frac{x}{2}-\frac{1}{2}\log|\cos x-\sin x|+C$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
A. $\int\frac{\sin x}{1+\cos x}dx$
let $y=1+\cos x⇒dy=-\sin x\,dx$
$⇒\int-\frac{dy}{y}=-\log|1+\cos x|+C$
B. $\int\frac{1}{1-\tan x}dx⇒\int\frac{1}{1-\frac{\sin x}{\cos x}}dx$
$⇒\int\frac{\cos x}{\cos x-\sin x}dx⇒\frac{1}{2}\frac{2\cos x}{\cos x-\sin x}dx$
$⇒\frac{1}{2}\int\frac{\cos x-\sin x}{\cos x-\sin x}+\int\frac{\cos x+\sin x}{\cos x-\sin x}dx$
$=\frac{x}{2}+\frac{\log|\cos x-\sin x|+C}{2}$
C. $\int\frac{e^{\tan^{-1}x}}{1+x^2}dx$
Let $y=\tan^{-1}x⇒dy=\frac{dx}{1+x^2}$
$⇒\int e^ydy=e^y+c=e^{\tan^{-1}x}+c$
D. $\int\frac{1}{x+x\log x}dx$
Let $y=1+\log x⇒dy=\frac{1}{x}dx$
$⇒\frac{dy}{y}=\log y +c=\log(\log x+1)+c$