IIT JEE , AIEEE Forum - Mathematics , Integral Calculus - Very Good Integration Question(rates to all correct replys)

Anupamagrawal13a

Ques-_{[ 0]}

^{[pie/6 ]} ((

3cos2x - 1)/cosx)dx

Answer will be posted after 5 or more Replys.

Anupamagrawal13a

Please someone solve this.

sameerh522

sin^-1^{[ ]}

2(2-^{[ ]}

3)/ 3^{[ ]}

3} -1/3

sameerh522

2pi/90 - 1/3

abhujabal

Hey my answer comes out to be

3 - (1+

3)ln

3

Please rate if it's correct......

astrojith

My answer is :

3 - log

3

Please rate if it is correct. :)

richa_dpsvk

abhujabal is correct...the answer is

3-(1+

3)ln

3

abhujabal

hey thanx richa
plz someone rate ....
where is anupamagarwal she said to rate the correct repies.....

astrojith

abhu ma' man, could you please post the idea for solving it too ? Thats the point of being here, right ? Sharing knowledge. :)

abhujabal

OK my friends I am giving the solution First break cos2x as 2cos²x - 1

We get

_{[ ]}

^{[ ]} (2

3cos²x-

3-1)/cosx dx

= _{[ ]}

^{[ ]} 2

3cosx dx - (

3+1)_{[ ]}

^{[ ]} secx dx

=2

3sinx |₀^/6- (1+

3)ln(secx + tanx) |₀^/6

⁼

3 - (1+

3)ln

3

I hope now u r satisfied........

I hopw I can expect some points now......

k_greatgirl

hi ma answer is

3-(1+

3)ln

3

plz rate if correct ok BYE
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