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

evaluate the following integral ∫ 1 / cos 3x – cos x dx

evaluate the following integral
∫ 1 / cos 3x – cos x dx

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans:
Hello student,
Please find the answer to your question below

I = \int \frac{1}{cos3x-cosx}dx
cos3x - cosx = 2sin(2x)sin(-x)
cos3x - cosx = -2sin(2x)sin(x)
I = \int \frac{-1}{2sin2xsinx}dx
I = \int \frac{-1}{2.2sinxcosxsinx}dx
I = \int \frac{-1}{4sin^{2}xcosx}dx
I = \int \frac{-cosx}{4sin^{2}xcos^{2}x}dx
sinx = t
cosxdx = dt
I = \int \frac{-1}{4t^{2}(1-t^{2})}dt
I = -\int \frac{1-t^{2}+t^{2}}{4t^{2}(1-t^{2})}dt
I = -\int \frac{1}{4t^2}dt-\int\frac{1}{4(1-t^{2})}dt
I = \frac{1}{4t}-\frac{1}{4}log(\frac{1+t}{1-t})+c
I = \frac{1}{4sinx}-\frac{1}{4}log(\frac{1+sinx}{1-sinx})+c

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free