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Question

Question: If \(\int_{}^{}\frac{\cos 4x + 1}{\cot x–\tan x}\)dx = k cos 4x + c, then...

If cos4x+1cotxtanx\int_{}^{}\frac{\cos 4x + 1}{\cot x–\tan x}dx = k cos 4x + c, then

A

k = – ½

B

k = – 1/8

C

k = – ¼

D

None of these

Answer

k = – 1/8

Explanation

Solution

=(2x+4)x2+4x+3dxx2+4x+3dx= \int ( 2 x + 4 ) \sqrt { x ^ { 2 } + 4 x + 3 } d x - \int \sqrt { x ^ { 2 } + 4 x + 3 } d x. cos x. sin x dx

= =I1I2,(say)= I _ { 1 } - I _ { 2 } , ( s a y )sin 2x dx = – 18\frac{1}{8}cos4x + c \ k = – 1/8