Messages posted by allamraju

2)For a point (x0,y0) to lie in the interior of S,the conditon is S00<0.

So,In this case,S00=4+64-4+32-p<0

96-p<0 (or) p>96.

Now,Let

Now,put 2x=t so that it becomes

So, and hence the given integral is

You made a mistake here.

-[x2-(a+b)x+ab]=(a+b/2)2-ab-[x-(a+b/2)]2.

Plz check it,U forgot the ab term.

Put a+bx=t then bdx=dt.So,the integral becomes

Since ,we have,

=

See this.I answered it here.

Put x=cos2k then dx=-sin2kdk.So,the integral becomes

Where .

First put x=t6 then dx=6t5dt.Then the integral becomes,

Let S=c+1+x2+x4+....+x2n+....upto infinite terms.

Then S'=2x+4x3+....+2nx2n-1+...upto infinite terms

S'=(2x+4x3+...+2nx2n-1)=f(x)

Integral of S'=S

where S=c+1+x2+....+infinite terms=c+1/1-x2 since 0<x<1.

Thus,

Simple yaar,Since G is the centroid,

G=(-6+a+c/3,0+b+d/3)=(-2,-2)

a+c-6=-6 and b+d=-6 (or) a+c=0,b+d=-6.

Firstly,1+cos3x=4cos3x-3cosx+1

=[4cos3x-2cos2x]+[2cos2x-3cosx+1]

=2cos2x(2cosx-1)+(2cosx-1)(cosx-1)

=(2cosx-1)[2cos2x+cosx-1]

= (2cosx-1)(cosx+cos2x).

So,

Let given vertex be A=(-6,0) and the other vertices be B=(a,b) and C=(c,d).

G=(-2,-2) gives a+c=0 and b+d=-6.

Slope of AB=b/a+6 and slope of AC=d/c+6.

So,eqn. of altitudes perpendicular to AC and AB are

(a+6)x+by=(a+6)c+bd and (c+6)x+dy=(c+6)a+bd

Both these intersect at (0,0) and so,ac+bd+6c=0=ac+bd+6a.

This gives a=c and so,a=c=0 since a+c=0.

So,bd=0b=0 or d=0.If b=0 then d=-6 and viceversa.

Hence the other two vertices are (0,0) and (0,-6).

Let us put then we have

So the integral becomes

1)f(1/x)=

Put k=1/t then dt=(-1/k2)dk.So,the integral becomes

So,f(x)+f(1/x)=

Putting x=e,we get f(e)+f(1/e)=1/2.

[1+sinpix]=0 for x in (-1/2,0) and

[1+sinpix]=1 for x in (0,1/2)

So,the integral becomes

Now,If 0<x<1/2 then 1<x2<5/4 and so,[x2+1]=1.Thus,the answer is

=1/2+1/2=1.

Put x=t2 then dx=2tdt.So,the integral becomes

Now put tanx=t then the integral becomes

whose integral is

whose integral is ln(ex+e-x)+c.

Simple man! where c is determined using that f(2)=0 which gives c=-(16+1/8)=-129/8.

So,the answer is

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2)For a point (x₀,y₀) to lie in the interior of S,the conditon is S₀₀<0.

So,In this case,S₀₀=4+64-4+32-p<0

-[x²-(a+b)x+ab]=(a+b/2)²-ab-[x-(a+b/2)]².

Put x=cos²k then dx=-sin2kdk.So,the integral becomes

First put x=t⁶ then dx=6t⁵dt.Then the integral becomes,

Let S=c+1+x²+x⁴+....+x²ⁿ+....upto infinite terms.

Then S'=2x+4x³+....+2nx^2n-1+...upto infinite terms

S'=(2x+4x³+...+2nx^2n-1)=f(x)

where S=c+1+x²+....+infinite terms=c+1/1-x² since 0<x<1.

Firstly,1+cos3x=4cos³x-3cosx+1

=[4cos³x-2cos²x]+[2cos²x-3cosx+1]

=2cos²x(2cosx-1)+(2cosx-1)(cosx-1)

=(2cosx-1)[2cos²x+cosx-1]

Put k=1/t then dt=(-1/k²)dk.So,the integral becomes

Now,If 0<x<1/2 then 1<x²<5/4 and so,[x²+1]=1.Thus,the answer is

Put x=t² then dx=2tdt.So,the integral becomes

whose integral is ln(e^x+e^-x)+c.