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  • 1 answers

Aditya Roy 6 years, 10 months ago

1/sinxsina dx =1/sina int 1/sinxdx =1/sina int cosecxdx =1/sina[ log(cosecx+cotx)]+c
  • 1 answers

Asit Bafna 6 years, 10 months ago

Ans=7
  • 1 answers

Amit Chauhan 6 years, 10 months ago

Dy/dx=cosx/3 - 2/3
  • 3 answers

Aditya Roy 6 years, 10 months ago

Haa basically zisme gp lgta jai

Amit Chauhan 6 years, 10 months ago

Sb important h ncert m h jo

Akshat Tiwari 6 years, 10 months ago

No,not imp.
  • 3 answers

Amit Chauhan 6 years, 10 months ago

Cos(cosx)

Hardik Omre 6 years, 10 months ago

Let cosx=t diff. both side w. r. t then -sinx=dt/dx then dx=-dt/sinx then inegration of Sinxsin(cos)= -cos(t)then -cos(cos) is ans.

Kartik Periwal 6 years, 10 months ago

-cos(cosx)
  • 1 answers

Amit Chauhan 6 years, 10 months ago

1/xlogx+1/x
  • 1 answers

Akshat Tiwari 6 years, 10 months ago

City in new jersey
  • 1 answers

Saket Kumar Suman 6 years, 10 months ago

By outline, in sphere (rad.=r) draw triangle(cone∆) pts. A,B,C ,mark centre O in it,OC =R,<COD=2<CAD <CAD=@, <COD=2@ the height h (AD)has 2parts (r+rcos2@)=r(1+cos2@)=2rcos^2(@) and CD=rsin2@ (v)Vol.=1/3πr^2h =1/3π(CD^2)AD =1/3π(r^2 sin^2(2@)) * 2rcos^2(@) =2/3π r^3 {4 sin^2(@) cos^2(@)} cos^2(@) =8/3π r^3 {sin^2(@) cos^4(@)} Now diffn. In terms of @ Apply product rule ( Take, 8/3πr^3= ₹) dv/d@ =₹{2sin@cos^5(@)-4cos^3(@)sin^3(@)} =2₹sin@cos^3(@) {cos^2(@)-2sin^2(@)} =0(max.) ={cos^2(@)-2sin^2(@)}=0 ={1- sin^2(@)-2sin^2(@)}=0 =1-3sin^2(@)=0 =sin^2(@)=1/3 Now h AD=r(1+cos2@)=r{1+ 1- 2sin^2(@)} =r{2-2*1/3} =r{2-2/3} =12*4/3 =16 cm
  • 1 answers

Nitin Gole 6 years, 10 months ago

Originally Answered: Is integral of mod of X possible? We first note that: ddx(|x|)=|x|xddx(|x|)=|x|x where x cannot take the value 0. Hence, I=∫|x|dxI=∫|x|dx =∫|x|xxdx=∫|x|xxdx =∫xd(|x|)=∫xd(|x|) =x|x|−∫|x|dx=x|x|−∫|x|dx =x|x|−I=x|x|−I ⟹2I=x|x|⟹2I=x|x| ⟹I=x|x|2+C
  • 2 answers

Preeti Takhar 6 years, 10 months ago

It will form identity of 2tan^-1x ans will be 2/1+a^2x

Preeti Takhar 6 years, 10 months ago

Differentiate krna h kya
  • 1 answers

Preeti Takhar 6 years, 10 months ago

Log (x/a+bx)^x-1(2bx+a/a+bx)+log (x/a+bx)^x ×log (logx/a+bx). May be
  • 2 answers

Akash Porwal 6 years, 10 months ago

just replace 3x = t then use identity of cos2● and solve at last u will get answer 1/3 tan3x/2

Akash Porwal 6 years, 10 months ago

very simple
  • 1 answers

Anjana Sherawat 6 years, 10 months ago

3sinx -4sin ^3x
Q2
  • 0 answers
  • 1 answers

Akash Porwal 6 years, 10 months ago

May not be but you must knw how these formulas came
  • 4 answers

Preeti Takhar 6 years, 10 months ago

At least 8 hours a dayy

Dharshika As 6 years, 10 months ago

100% sure that if you study properly without wasting your time then you can get above 70% easily proper management time is esswntial

Rupas Chettri 6 years, 10 months ago

How much we should study

Deepak Jindal 6 years, 10 months ago

Yess
  • 3 answers

Harjeet Singh 6 years, 10 months ago

Solbe it by epdx by comparing with dy/dx+py=q where u geg p=2tanx and q =sinx after that solve it

Akash Porwal 6 years, 10 months ago

cosx-2cos^2x

Aman Goyal 6 years, 10 months ago

Solve by integrating facter
  • 0 answers
  • 0 answers

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