CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Solve:
x2x2a2dx

Open in App
Solution

x2x2a2dxLetx=asinθ1=acosθdθdxdx=acosθdθ(1)x2x2a2=a2sin2θa2sin2θa2=tan2θ(1sin2θ=cos2θandsinθcosθ=tanθ)x2x2a2dx=(tan2θ).acosθdθ=a(sec2θ1)cosθdθ(1+tan2θ=sec2θ)=a(cosθsecθ)dθ=asinθlog|secθ+tanθ|+C(secθ=log|secθ+tanθ|)x=asinθsinθ=xax2x2a2dx=a.xalogaa2x2+xa2x2+C(fromthefiguregetthevalueofsecθandtanθ)=xloga+xa2x2+C
1096472_1092217_ans_3b8c9c6657b542b9810de170e9fc782e.jpg

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