wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If 10dtt2+2tcosα+1x233t2sin2tt2+1dtx2=0, (0<α<π), then the value of x is

A
±α2sinα
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
±2sinαα
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
±αsinα
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
±2sinαα
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D ±2sinαα
10dtt2+2tcosα+1x233t2sin2tt2+1dtx2=0
We know,
aaf(x)dx=0 if f(x)+f(x)=0
Let,
I=10dtt2+2tcosα+1I=10dt(t+cosα)2+sin2αI=1sinα[tan1(t+cosαsinα)]10I=1sinα[tan1cotα2tan1cotα]I=1sinα[tan1tan(π2α2)tan1tan(π2α)]I=1sinα[(π2α2)(π2α)]I=α2sinα

From the equation, we get
Ix22=0x2=4sinααx=±2sinαα

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Integration by Partial Fractions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon