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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

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

Grade:11

4 Answers

Akash Kumar Dutta
98 Points
8 years ago

in the third put x+a = t ...x=t-a
dx=dt
the 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

Akash Kumar Dutta
98 Points
8 years ago

for the first,
we know 1+cosx=2cos ^2 x/2
hence 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

Akash Kumar Dutta
98 Points
8 years ago

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,
tan
θ=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)

Akash Kumar Dutta
98 Points
8 years ago

Sorry man not to able to solve thos one....but i think we need to proceed like this
these 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 = u
then by parts try to integrate
hope it helps...

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