Click to Chat

1800-2000-838

+91-120-4616500

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
Chilukuri Sai Kartik Grade: 12
```        1.∫(Sin4x/Cos2x).dx=?
2.∫{1+sin(x/2)}1/2 dx=?
Please give elaborate answers for these at the earliest...```
6 years ago

## Answers : (2)

510 Points
```										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
```
6 years ago
Aman Bansal
592 Points
```										dear chilukuri,
1.∫(Sin4x/Cos2x).dx
Possible intermediate steps:integral sec(2 x) sin(4 x) dxSimplify the integrand sin(4 x) sec(2 x) to get 4 sin(x) cos(x):= integral 4 sin(x) cos(x) dxFactor out constants:= 4 integral sin(x) cos(x) dxFor the integrand sin(x) cos(x), substitute u = cos(x) and du = -sin(x) dx:= -4 integral u duThe integral of u is u^2/2:= -2 u^2+constantSubstitute 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])
We are all IITians and here to help you in your IIT JEE preparation.
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.
AMAN BANSAL

```
6 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

## Other Related Questions on Integral Calculus

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details