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Evaluate the following integrals:-
Q-1)∫1/√1+cosx dx
Q-2)∫{1+TANXTAN(X+θ)} dx
Q-3)∫sin(x-α)/sin(x+α) dx
Q-4)∫secxtanx/3x+5 dx
in the third put x+a = t ...x=t-adx=dtthe eq becomes. int. [sin (t-2a)/sint] dt int. [cos 2a - cot t.sin2a ] dt =cos 2a.t - sin 2a.log |sin t| + C = (x+a)cos 2a - sin 2a.log |sin (x+a) | +C...ANS
for the first,we know 1+cosx=2cos ^2 x/2hence the eq. becomes,int [ 1/(root2)cos (x/2)]dx= 1/root2 . int [ 1/cos (x/2)]dx...now put x/2=m....such that dx=2dm= 1/root2 .int [2secm]dm=(root 2).log |sec m + tan m | + C= (root2).log | sec x/2 + tan x/2 | +C...ANS
for the SECOND...i had difficulty using theta here so in the midway i have taken (o) as theta....so dont get confused!!!we know,tanθ=tan(x+θ-x)=tan(x+θ)-tanx/ 1+tan x.tan(n+θ)hence 1 + tanxtan(x+θ)=1/tan (o) [ tan(x+(o) ) - tanx]hence the eq,. becomes,1/tan(o) int [ tan( x+(o) ) - tanx ]1/tan (o) [ -log |cos ( x+(o) ) + log|cosx| ]1/tan(o) [ log |cos x / cos ( x+(o) ) ]= log | cos x / cos ( x + (o) ) | ______________________ + C ........ANS tan(o)
Sorry man not to able to solve thos one....but i think we need to proceed like thisthese types of sums are impossible without int by parts.....we know the integral of secxtanx=secx so take it as v,and 1/3x+5 = uthen by parts try to integratehope it helps...
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