integration of dx/(3+2sinx-cosx)
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the given integral is;
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
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
karthik4297:
what did u substituted ?
pleae clear it.
ok
ur solution is not clear..plz clear it...or u can send an image..
ur solution is not clear..plz clear it...or u can send an image..
ok man.......
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