0

Guest

Courses New

integrate cosec(x-3T)cosec(x-2T).dT limit 0 to 4T

integrate cosec(x-3T)cosec(x-2T).dT limit 0 to 4T

Grade:

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
I = \int_{0}^{4T}cosec(x-3T).cosec(x-2T)dx
I = \int_{0}^{4T}\frac{1}{sin(x-3T).sin(x-2T)}dx
I = \frac{1}{sin(T)}\int_{0}^{4T}\frac{sin(T)}{sin(x-3T).sin(x-2T)}dx
I = \frac{1}{sin(T)}\int_{0}^{4T}\frac{sin[(x-2T)-(x-3T)]}{sin(x-3T).sin(x-2T)}dx
I = \frac{1}{sin(T)}\int_{0}^{4T}\frac{sin(x-2T).cos(x-3T)-cos(x-2T).sin(x-3T)}{sin(x-3T).sin(x-2T)}dx
I = \frac{1}{sin(T)}\int_{0}^{4T}(\frac{cos(x-3T)}{sin(x-3T)}-\frac{cos(x-2T)}{sin(x-2T)})dx
I = \frac{1}{sin(T)}(ln(sin(x-3T))-ln(sin(x-2T)))_{0}^{4T}
I = \frac{1}{sin(T)}(ln(\frac{sin(x-3T)}{sin(x-2T)}))_{0}^{4T}I = \frac{1}{sin(T)}(ln(\frac{sin(T)}{sin(2T)})-ln(\frac{sin(3T)}{sin(2T)}))
I = \frac{1}{sin(T)}.ln(\frac{sin(T)}{sin(3T)})
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

Think You Can Provide A Better Answer ?

Other Related Questions on Integral Calculus

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More