Q. 544.5( 2 Votes )

# Evaluate the foll

Let

As we have the trigonometric identity

to evaluate this integral we use x2 = a2cos 2θ

2xdx = –2a2sin(2θ)dθ (Differentiating both sides)

xdx = –a2sin(2θ)dθ

When x = a, a2cos 2θ = a2 cos 2θ = 1

2θ = 0 θ = 0

So, the new limits are and 0.

Also,

Substituting this in the original integral,

But, we have 2 sin2θ = 1 – cos 2θ

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

Evaluate the follMathematics - Exemplar

Evaluate: <a nameMathematics - Board Papers

If <img widMathematics - Board Papers

<span lang="EN-USRD Sharma - Volume 2

Evaluate the follMathematics - Exemplar

<span lang="EN-USRD Sharma - Volume 2

<span lang="EN-USRD Sharma - Volume 2

<span lang="EN-USRD Sharma - Volume 2

Evaluate : <span Mathematics - Board Papers

<span lang="EN-USRD Sharma - Volume 2