Question

X.COS²X dx ka integration ?

Answer 1

`I=1/4x^2+1/4xsin(2x)+1/8cos(2x)+C`

Explanation:

We want to solve

`I=intxcos^2(x)dx`

Using the trig identity `color(red)(cos^2(x)=1/2+1/2cos(2x)`

`I=intx(1/2+1/2cos(2x))dx`

`color(white)(I)=1/2intxdx+intxcos(2x)/2dx`

The first integral is trivial, for the second apply IBP

`color(blue)(intudv=uv-intvdu)`

Let `color(blue)(u=x=>du=dx`

And `color(blue)(dv=cos(2x)/2dx=>v=sin(2x)/4`

`I_1=intxcos(2x)/2dx`

`color(white)(I_1)=xsin(2x)/4-intsin(2x)/4dx`

`color(white)(I_1)=xsin(2x)/4+cos(2x)/8+C_1`

Thus

`I=1/4x^2+1/4xsin(2x)+1/8cos(2x)+C`

Answer 2

`I=1/8[2x^2+2sin2x+cos2x]+C`

Explanation:

Here,

`I=intxcos^2xdx`

`=intx((1+cos2x)/2)dx`

`=1/2intxdx+1/2intxcos2xdx`

`I=1/2*x^2/2+1/2I_1...to(A)`

Where, `I_1=intxcos2xdx`

`"Using"color(blue)" Integration by Parts:"`

`int(uv)dx=uintvdx-int(u'intvdx)dx`

Let, `u=x and v=cos2x=>u'=1andintvdx=(sin2x)/2`

So,

`I_1=x*(sin2x)/2-int1*(sin2x)/2dx`

`=x/2sin2x-1/2intsin2xdx`

`=x/2sin2x-1/2((-cos2x)/2)+c`

`:.I_1=x/2sin2x+(cos2x)/4+c`

Hence, from `(A)`

`I=1/2*x^2/2+1/2[x/2sin2x+(cos2x)/4+c]`

`=x^2/4+x/4sin2x+1/8cos2x+c/2`

`I=1/8[2x^2+2sin2x+cos2x]+C,where.C=c/2`