wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

x·dydx·tanyx=ytanyx+x, y12=π6. Area bounded by x=0,x=12,y=yx


Open in App
Solution

Step 1: Note the given data

y12=π6

The equation is x·dydx·tanyx=ytanyx+x

Dividing the equation by x

dydx·tanyx=yxtanyx+1........1

Let y=vx..........2

Differentiating with respect to x

dydx=v+xdvdx............3

Step 2: Substituting equations (2) and (3) in equation (1)

v+xdvdx·tanv=vtanv+1

vtanv+xtanvdvdx=vtanv+1

tanvdv=1xdx

Integrating both sides,

tanvdv=1xdx

-lncosv=lnx+lnc

-cosv=xc

c=-cosvx

c=-cosyxx

yx=cos-1-cx

'yx=cos-1cxcos-1-x=cos-1x

y=xcos-1cx

Given that y12=π6

π6=12cos-112c

cosπ3=12c

12=12c

c=1

y=xcos-1x

Step 3: Find the area for the given boundary

Integration formula:

  • uvdx=vudx-dvdxudxdx
  • xndx=xn+1n+1+C

Derivative formula

dcos-1xdx=-11-x2

A=012xcos-1xdx

A=cos-1x·x22012-012-11-x2x22dx

A=12π4·12+12012x21-x2dx

A=π16+12012x21-x2dx

Let x=cosθ;dx=-sinθdθ

x120
θπ4π2

A=π16+12π2π4-cos2θsinθ1-cos2θdθ

A=π16-12π2π4cos2θsinθsinθdθ

A=π16-12π2π4cos2θdθ

A=π16-12π2π4cos2θ+12dθcos2θ=2cos2θ-1

A=π16-14π2π4cos2θ+1dθ

A=π16-14sin2θ2+θπ2π4

A=π16-1412+π4-0+π2

A=π16-142+π-2π4

A=π16-2-π16

A=2π-216

A=π-18sq.unit

Hence the required area is π-18sq.unit.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Trigonometric Functions in a Unit Circle
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon