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## Answers

I=∫dx3+2sinx+cosx

substitute sinx=2tanx21+tan2x2 and cosx=1−tan2x21+tan2x2

I=∫dx3+2*2tan(x/2)1+tan2(x/2)+1−tan2(x/2)1+tan2(x/2)=∫1+tan2(x/2)3+3tan2(x/2)+4tan(x/2)+1−tan2(x/2)dx=∫sec2(x/2)2tan2(x/2)+4tan(x/2)+4dx=∫sec2(x/2)2[{tan(x/2)+1}2+1]dx let tan⎛⎝⎜⎜⎜x/2⎞⎠⎟⎟⎟=t⇒12sec2⎛⎝⎜⎜⎜x/2⎞⎠⎟⎟⎟dx=dt

I=∫1(t+1)2+1dtI=tan−1(t+1)+CI=tan−1{tanx2+1}+C

where C is an integration constant.

hope this helps you