∫11+cosx+sinxdx=
14log1+tanx2+c
12log1+tanx2+c
2log1+tanx2+c
12log1-tanx2+c
Explanation for the correct answer:
Integrating the given integral:
Let I=∫11+cosx+sinxdx
We know that cosx=1-tan2x21+tan2x2,sinx=2tanx21+tan2x2
I=∫11+1-tan2x21+tan2x2+2tanx21+tan2x2dx=∫sec2x221+tanx2dx
Substituting
tanx2=tDifferentiating12sec2x2dx=dt
Therefore,
I=∫dt4(1+t)=14log|1+t|+c=14log1+tanx2+c
Hence, ∫11+cosx+sinxdx=(14)log|1+tan(x2)|+c
Therefore, the correct answer is option (A).
loge(n+1)−loge(n−1)=4a[(1n)+(13n3)+(15n5)+...∞] Find 8a.