1.primitive of (3x^4-1)/(x^4+x+1)with respect to x. 2.integral of (1+2cot(cotx+cosecx)^-1/2
Nikhil Raj , 7 Years ago
Grade 12
Integral Calculus> 1.primitive of (3x^4-1)/(x^4+x+1) with re...
4 AnswersRushang Patel
Last Activity: 6 Years ago
Dear Student,
Let I = ∫{1 + 2 cot x(cot x + cosec x)}^1/2dx
∫{1 + 2 cot^2x + 2 cot x*cosec x}^1/2dx
∫{1 + cot^2x + cot^2x + 2 cotx*cosec x}^1/2 dx
∫{cosec^2x + cot^2x + 2 cosec x *cotx}^1/2dx
∫{(cosec x + cot x)^2}^1/2dx
∫(cosec x + cot x) dx
log|cosec x − cot x| + log|sin x| + C'
log|(cosecx-cotx)sinx|+C'
log|cosecx*sinx - cotx*sinx| +C'
log|1-cosx| +C'
log|2sin^2x/2| + C'
log2 + 2log|(sinx/2)| +C'
Now log2 +C'=C
I = 2log|sinx/2| + C is the ANSWER
Regards
Rushang Patel
Last Activity: 6 Years ago
Let I = ∫{1 + 2 cot x(cot x + cosec x)}^1/2dx
∫{1 + 2 cot^2x + 2 cot x*cosec x}^1/2dx
∫{1 + cot^2x + cot^2x + 2 cotx*cosec x}^1/2 dx
∫{cosec^2x + cot^2x + 2 cosec x *cotx}^1/2dx
∫{(cosec x + cot x)^2}^1/2dx
∫(cosec x + cot x) dx
log|cosec x − cot x| + log|sin x| + C'
log|(cosecx-cotx)sinx|+C'
log|cosecx*sinx - cotx*sinx| +C'
log|1-cosx| +C'
log|2sin^2x/2| + C'
log2 + 2log|(sinx/2)| +C'
Now log2 +C'=C
I = 2log|sinx/2| + C is the ANSWER
Regards
Rushang Patel
Last Activity: 6 Years ago
Let I = ∫{1 + 2 cot x(cot x + cosec x)}^1/2dx
∫{1 + 2 cot^2x + 2 cot x*cosec x}^1/2dx
∫{1 + cot^2x + cot^2x + 2 cotx*cosec x}^1/2 dx
∫{cosec^2x + cot^2x + 2 cosec x *cotx}^1/2dx
∫{(cosec x + cot x)^2}^1/2dx
∫(cosec x + cot x) dx
log|cosec x − cot x| + log|sin x| + C'
log|(cosecx-cotx)sinx|+C'
log|cosecx*sinx - cotx*sinx| +C'
log|1-cosx| +C'
log|2sin^2x/2| + C'
log2 + 2log|(sinx/2)| +C'
Now log2 +C'=C
I = 2log|sinx/2| + C is the ANSWER
Rushang Patel
Last Activity: 6 Years ago
Answer no 2
Let I = ∫{1 + 2 cot x(cot x + cosec x)}^1/2dx
∫{1 + 2 cot^2x + 2 cot x*cosec x}^1/2dx
∫{1 + cot^2x + cot^2x + 2 cotx*cosec x}^1/2 dx
∫{cosec^2x + cot^2x + 2 cosec x *cotx}^1/2dx
∫{(cosec x + cot x)^2}^1/2dx
∫(cosec x + cot x) dx
log|cosec x − cot x| + log|sin x| + C'
log|(cosecx-cotx)sinx|+C'
log|cosecx*sinx - cotx*sinx| +C'
log|1-cosx| +C'
log|2sin^2x/2| + C'
log2 + 2log|(sinx/2)| +C'
Now log2 +C'=C
I = 2log|sinx/2| + C is the ANSWER
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