Question

If `int(cos4x+1)/(cotx-tanx)dx=kcos4x+c`, then k= (A) `-1/4` (B) `-1/2` (C) `-1/8` (D) none of these
A. `A=(1)/(8),BinR`
B. `A=-(1)/(8),B inR`
C. `A=(1)/(4),BinR`
D. none of these

Answer 1

Correct Answer - b
We have , `I=int(1+cos 4x)/(cotx-tanx)dx`
`rArrI=int(2cos^(2)2x sinx cosx)/(cos^(2)x-sin^(2)x)=intsin2x cos2x dx`
`rArrI=(1)/(2)intsin 4x dx =-(1)/(8)cos4x +B`
Hence , `A=-(1)/(8)andBinR`