badge image

Enroll For Free Now & Improve Your Performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 12

                        

How do you integrate y = e^x sin2x dx?

5 years ago

Answers : (1)

grenade
2061 Points
							
Let: 

u = sin(2x) 
du = 2cos(2x) dx 
dv = e^x dx 
v = e^x 

Then: 

L = uv - ∫ v du 
==> L = e^x*sin(2x) - 2 ∫ e^x*cos(2x) dx 

Let: 

u = cos(2x) 
du = -2sin(2x) 
dv = e^x dx 
v = e^x 

By another round of integration by parts: 

L = e^x*sin(2x) - 2(uv - ∫ v du) 
==> L = e^x*sin(2x) - 2[e^x*cos(2x) + 2 ∫ e^x*sin(2x) dx] 
==> L = e^x*sin(2x) - 2e^x*cos(2x) - 4 ∫ e^x*sin(2x) dx 
==> L = e^x*sin(2x) - 2e^x*cos(2x) - 4L (since L = ∫ e^x*sin(2x) dx) 
==> 5L = e^x*sin(2x) - 2e^x*cos(2x) 
==> 5L = e^x * [sin(2x) - 2cos(2x)] 
==> L = e^x * [sin(2x) - 2cos(2x)]/5 
==> ∫ e^x*sin(2x) dx = e^x * [sin(2x) - 2cos(2x)]/5 + C 

Therefore, ∫ e^x*sin(2x) dx = e^x * [sin(2x) - 2cos(2x)]/5 + C. 
 
 

 approve if useful
5 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 51 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details