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Evaluate the following integrals : (i) int(cosx)/(2sinx+3cosx)dx (ii) int(2sinx+cosx)/(7sinx-5cosx)dx ,br. (iii) int(5sinx+6)/(2cosx+sinx+3)dx |
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Answer» Solution :`(i)` We have `cosx=K(2sinx+3cosx)+m(2cosx-3sinx)` `impliescosx=(2k-3m)sinx+(3k+2m)cosx` Equating the COEFFICIENTS `sinx` and `cosx` from both sides `3k+2m=1` `2k-3m=0` when `k=(3)/(13)` and `m=(2)/(13)` Thus `int(cosx)/(2sinx+3cosx)DX=(3)/(13)intdx+(2)/(13)int((2cosx-3sinx))/(2sinx+3cosx)dx` `=(3)/(13)x+(2)/(13)ln|2sinx+3cosx|` `(ii)` Let `2sinx+cosx=k(7sinx-5cosx)+m(7cosx+5sinx)` `implies2sinx+cosx=(7k+5m)sinx+(7m-5k)cosx` `implies7k+5m=2` and `7m-5k=1` when `m=(17)/(74)` and `k=(9)/(14)` Thus `int(2sinx+cosx)/(7sinx-5cosx)dx=(9)/(74)intdx+(17)/(74)int(7cosx+5sinx)/(7sinx-5cosx)dx` `=(9)/(14)x+(17)/(74)ln|7sinx-5cosx|+C` `(iii)` Let`5sinx+6=k(2cosx+sinx+3)+m(-2sinx+cosx)+n` `implies5sinx+6=(2k+m)cosx+(k-2m)sinx+3k+n` `implies2k+m=0`, `k-2m=5`, `3k+n=6` `impliesk=1`, `n=3`, `m=-2` Thus `int(5sinx+6)/(2cosx+sinx+3)dx=intdx-2int((cosx-2sinx))/(2cosx+sinx+3)dx+3int(1)/(2cosx+sinx+3)dx` `=x-2ln|2cosx+sinx+3|+3l_(1)`, say `l_(1)=int(1)/((2cosx+sinx+3)dx=int(1)/(2((1-tan^(2).(x)/(2))/(1+tan^(2).(x)/(2)))+(2tan.(x)/(2))/(1+tan^(2).(x)/(2))+3)dx` `=int(1+tan^(2).(x)/(2))/(2-2tan^(2).(x)/(2)+2tan.(x)/(2)+3+3tan^(2).(x)/(2))dx` `=int(sec^(2).(x)/(2))/(5+2tan.(x)/(2)+tan^(2).(x)/(2))dx` Let us put `tan.(x)/(2)=u` so that `sec^(2).(x)/(2)dx=2du` `=int(2)/(4+(1+u)^(2))du` `=2*(1)/(2)tan^(-1)((1+u)/(2))+C` `=tan^(-1)((tan.(x)/(2)+1)/(2))+C` Hence the given integral is `int(5sinx+6)/(2cosx+sinx+3)dx=x-2ln|2cosx+sinx+3|+3tan^(-1)((sin.(x)/(2)+cos.(x)/(2))/(2cos.(x)/(2)))+C` |
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