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If int(cos4x+1)/(cotx-tanx)dx=kcos4x+c, ...

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

Text Solution

Verified by Experts

The correct Answer is:
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`
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