If the integral ∫5tan⁡xtan⁡x−2dx=x+alog⁡|sin⁡x−2cos⁡x|+∣K then a is equal to

I=∫5tan⁡xtan⁡x−2dx=∫5sin⁡xsin−2cos⁡xdxLet 5sin⁡x=A(sin⁡x−2cos⁡x)+B(cos⁡x+2sin⁡x) Equating coefficients of sin⁡x and cos⁡x, we get 5=A+2B and 0=−2A+B⇒ A=1,B=2∴ I=∫(sin⁡x−2cos⁡x)+2(cos⁡x+2sin⁡x)sin⁡x−2cos⁡xdx =x+2log⁡|sin⁡x−2cos⁡x|+K∴ a=2