CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

If the reduction formula for In=sin nxcos xdx is given by
In+In2=2cos{(n1)x}n1, then sin 3xcos xdx is.

A

cos 2x+log(sec x|+C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

cos 2xlog(sec x|+C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

cos 2xlog(sec x|+C
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

cos 2x+log(sec x|+C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C
cos 2xlog(sec x|+C
We know the reduction formula for In=sin nxcos xdx as :
In+In2=2cos{(n1)x}n1
Now, to find In=sin 3xcos xdx
we take n=3, thus substituting it back in the reduction formula we get
I3+I1=2cos 2x2,
Here, I1=sinxcosxdx=ln(secx|+C,
Thus we get I3=cos 2xln(secx|+C
So, Option c. is correct.

flag
Suggest Corrections
thumbs-up
0
BNAT
mid-banner-image