#### Thank you for registering.

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

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
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

# 1.∫(Sin 4 x/Cos 2 x).dx=? 2.∫{1+sin(x/2)} 1/2 dx=? Please give elaborate answers for these at the earliest... 10 years ago

I = (sinx/2 + 1)1/2 dx

(sinx/2+1) = (sinx/4)2 + (cosx/4)2 + 2cosx/4sinx/4   =  (sinx/4 + cosx/4)2

after putting this

I ={ sinx/4 + cosx/4 }dx

I= -4cosx/4 + 4sinx/4 + c

please approve my ans if u like it

10 years ago

dear chilukuri,

1.∫(Sin4x/Cos2x).dx

Possible intermediate steps:
integral sec(2 x) sin(4 x) dx
Simplify the integrand sin(4 x) sec(2 x) to get 4 sin(x) cos(x):
= integral 4 sin(x) cos(x) dx
Factor out constants:
= 4 integral sin(x) cos(x) dx
For the integrand sin(x) cos(x), substitute u = cos(x) and du = -sin(x) dx:
= -4 integral u du
The integral of u is u^2/2:
= -2 u^2+constant
Substitute back for u = cos(x):
= -2 cos^2(x)+constant

2.∫{1+sin(x/2)}1/2 dx

(4*(-Cos[x/4] + Sin[x/4])*Sqrt[1 + Sin[x/2]])/ (Cos[x/4] + Sin[x/4])

Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.