A) \[\frac{1}{2}+\frac{1}{2}\log |\cos x+\sin x|+C\]
B) \[\frac{x}{2}+\frac{1}{2}\log |\cos x-\sin x|+C\]
C) \[\frac{1}{2}+\frac{1}{2}\log |\cos x-\sin x|+C\]
D) \[\frac{x}{2}+\frac{1}{2}\log |\cos x+\sin x|+C\]
E) \[\frac{1}{2}+\frac{1}{2}\log |\cos x+\sin x|+C\]
Correct Answer: D
Solution :
\[\int{\frac{dx}{1+\tan x}}\] \[=\int{\frac{\cos x}{\sin x+\cos x}\times \frac{\cos x-\sin x}{\cos x-\sin x}}dx\] \[=\int{\frac{({{\cos }^{2}}x-\sin x.\cos x)}{({{\cos }^{2}}x-{{\sin }^{2}}x)}}dx\] \[=\frac{1}{2}\int{\frac{2{{\cos }^{2}}x}{\cos 2x}}dx-\frac{1}{2}\int{\frac{2\sin x.\cos x}{\cos 2x}}dx\] \[=\frac{1}{2}\int{\frac{(1+\cos 2x)}{\cos 2x}}dx-\frac{1}{2}\int{\frac{\sin 2x}{\cos 2x}}dx\] \[=\frac{1}{2}\int{\sec 2x\,}dx+\frac{1}{2}\int{dx-\frac{1}{2}}\int{\tan 2x\,}dx\] \[=\frac{1}{2}.\frac{1}{2}\log |\sec 2x+\tan 2x|\] \[+\frac{1}{2}.x-\frac{1}{2}.\frac{1}{2}\log |\sec 2x|+C\] \[=\frac{1}{4}\log |\sec 2x+\tan 2x|\] \[-\frac{1}{4}\log |\sec 2x|+\frac{x}{2}+C\] \[=\frac{x}{2}+\frac{1}{4}\log \left| \frac{\sec 2x+\tan 2x}{\sec 2x} \right|+C\] \[=\frac{x}{2}+\frac{1}{4}\log |1+\sin 2x|+C\] \[=\frac{x}{2}+\frac{1}{4}\log |({{\sin }^{2}}x+{{\cos }^{2}}x)\] \[+2\sin x.\cos x|+C\] \[=\frac{x}{2}+\frac{1}{4}\log |\sin x+\cos x{{)}^{2}}|+C\] \[=\frac{x}{2}+\frac{1}{2}\log |\cos x+\sin x)|+C\]
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