Solveeit Logo

Question

Mathematics Question on Integrals of Some Particular Functions

1sinxcosx\int\frac{1}{\sin x\, \cos x} dx is equal to

A

logtanx+C\log |\tan x| + C

B

logsin2x+C\log |\sin 2x| + C

C

logsecx+C\log |\sec x| + C

D

logcosx+C\log |\cos x| + C

Answer

logtanx+C\log |\tan x| + C

Explanation

Solution

Let I=(sin2x+cos2xsinxcosx)dxI=\int\left(\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}\right) d x
=(sin2xsinxcosx+cos2xsinxcosxdx)=\int\left(\frac{\sin ^{2} x}{\sin x \cos x}+\frac{\cos ^{2} x}{\sin x \cos x} d x\right)
=(tanx+cotx)dx=\int(\tan x+\cot x) d x
=logsecx+(logcosecx)+C= \log \sec x+(-\log \operatorname{cosec} x)+C
=logsecxcosecx+C=logtanx+C= \log \left|\frac{\sec x}{\operatorname{cosec} x}\right|+C=\log |\tan x|+C