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				   area under curve

4 years ago


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Are of the cure is the integral of the graph function

4 years ago

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please tell me the answers of the above 2 questions
the answer for sin^4(x) will be 3/8x-1/4sin2x+1/32sin4x+c
grenade 8 months ago
yhe answer for 1/cosx.dx wil be ln {tan mod (pie/4 + x/2) mod}+c
grenade 8 months ago
which 2 questions
grenade 8 months ago
simplify 2 nd term by partial fraction method Now above problem reduce to standard form and Do it yourself PLEASE APPROVE IF USEFUL
Aakash 6 months ago
hi there! i’m stuck n solving this one using the miracle substitution(weistrass).integral of 1/(sinx(3+2cosx). i am using this as this makes everything simple. but i dont know if there are...
hey take tan(x/2)=t, then sinx= 2t/(1+t^2) and cosx= (1-t^2)/ (1+t^2) and dx=2dt/(1+t^2), ur integral will become ∫ 2dt/(2t*(3+2(1-t^2)))= ∫ dt/(t(5-2t^2)) . This u can solve by partial...
Niranjan one month ago
“X” is wife of” Y” is it transtive?
yaaa!!! it is transitive
Gavvala Ganesh 5 months ago
RAKESH CHINDAM 5 months ago
SHANMUKESHWAR 5 months ago
f(0)=f(1)=0,f / (1)=2 &y=(e x )e f(x) dy/dx=
let, g’(0)=f’(1)e f(0) +f(1)e f(0) .f(0) =f’(1)+f’(0)=0+2=2
vivek 5 months ago
if cosx + cosy=4/5 and cosx - cosy=2/7 prove that 14tan(x-y/2)+5cot(x+y/2)=0
adding both you get cosx subtracting cosy so we also got sinx and siny if we use half angle formulas we can write sinx in terms of tanx/2 so we got that too so we can solve
girish 18 days ago
This question solve using half angle formulas we can write sinx in terms of tanx/2 so we got that then solve this problem.
Gowri sankar 18 days ago
Lab Bhattacharjee 18 days ago
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