Evaluate :∫[cosx-sinx][1+2sinxcosx]dx
-1(cosx–sinx)+c
[cosx+sinx][cosx–sinx]+c
-1(sinx+cosx)+c
-x(sinx+cosx)+c
Explanation for the correct option:
Finding ∫[cosx-sinx][1+2sinxcosx]dx:
Let I=∫[cosx-sinx][1+2sinxcosx]dx
We know that
1+2sinxcosx=sinx+cosx2
Substituting this identity we get
I=∫(cosx-sinx)(sinx+cosx)2dx
Substituting sinx+cosx=t
Differentiating this we get,
(cosx-sinx)dx=dt
Substituting the values of t and dt, we get
∴I=∫1t2dtI=−1t+c;∫xndx=xn+1n+1=−1sinx+cosx+c;[t=sinx+cosx]
Hence, option (C) is the correct answer.