Thank you for registering.

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

Please check your email for login details.
Close

MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

⇐ BackProceed to Checkout ⇒
There are no items in this cart.
Continue Shopping
Grade: 

                        
intefrated the function f(x) on R if f(x) =sinx.sin2x.sin3x.dx

							
intefrated the function f(x)  on R
if f(x) =sinx.sin2x.sin3x.dx 
5 months ago

Answers : (1)

Anand Kumar Pandey
askIITians Faculty
4324 Points 
							Dear Student

LetI=∫sinx.sin2x.sin3xdx
=(1/2) ∫2sinx.sin2x.sin3xdx
= (1/2) ​∫sinx.(2sin2x.sin3x)dx
We know that[2sinA.sinB=cos(A−B)−cos(A+B)]
So,= (1/2)∫sinx.(cosx−cos5x)dx
=(1/2)(1/2)∫2sinx.cosxdx−2×21​∫2sinx.cos5xdx
= (1/4) ​∫sin2xdx−(1/4)​∫(sin6x−sin4x)dx
=− (1/8)cos2x​+(1/24)cos6x​−(1/16)cos4x​+c

Thanks
4 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Other Related Questions on Integral Calculus

wvacescqwacswvacescqwacswvacescqwacswvacescqwacswvacescqwacswvacescqwacswvacescqwacswvacescqwacswvac... Answer & Earn Cool Goodies
{ ( cos ^-1x { √(1-x^2) }]^-1) ÷ ( log e { 1 + ( sin(2x√(1-x^2)) ÷ 2) }) dx Answer & Earn Cool Goodies
Çözümünü yaparmısınız rica etsem teşekürler şimdiden
In this particular question how would we think that we have to put t=tan(x/2)????
The ACS has launched a new program called Ask an ACS Chemist . This is a free utility available to...
View all Questions »


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