IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

barath90

Please help me to solve the attached problem

Thanks and regards,

Barath

joyfrancis

we have to find the value of a ? or do we have to prove this equation>?What is your question.?

aansh_c

i'll giv 2 answers

v hv 2 prove it rt

so dt means derivative of lhs is rhs rt?

proceed it directly by diffrentiating

or integrate d rhs

term 2 integrate

(a^2-x^2)^(1/2)

theta=@ ok?

put x=asin@...............................1)

dx=acos@d@

term 2 integrate

acos@*acos@*d@

(acos@)^2d@

a^2(1+cos2@)/2 * d@

(a^2)/2 * d@ + (a^2*cos2@)/2 * d@

(a^2*@)/2 + (a^2*sin@*cos@)/2

nw fm 1)

a^2*sin(inverse)(x/a) + (x*(a^2-x^2)^(1/2))/2

hope u r abl 2understnd it

puneet

hey

I hope that u have got an idea from the reply above .. in such questions always put the x = asin

sort of and u will see that it is all okay. Such are always famous questions so remember these tricks by heart .

cheers

iitkgp_bipin

This can be differentiated easily.

You must know the differentiation of standard terms and the product rule of differentiation. Arranging the terms will give you RHS. Still if you are not getting do report.

barath90

This is a differentiation problem. We have to prove that lhs = rhs. I am not getting it.

Sorry for not mentioning that.

Barath

karthik2007

this is a straight differentiation problem. Differentiate directily using chain rule and product rule, and you'll get the answer easily

cutepooja

ya karthik is correct

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