Question
∫π2−π2 sin2x dx=
Answer: Option B
:
B
∫π2−π2sin2xdx=2∫π20sin2xdx=2r(32)r(12)2r(2+22)=π2
Was this answer helpful ?
:
B
∫π2−π2sin2xdx=2∫π20sin2xdx=2r(32)r(12)2r(2+22)=π2
Was this answer helpful ?
More Questions on This Topic :
Question 3. ∫∞0 log(1+x2)1+x2dx=....
Question 5. In ∫baf(y) dy, what is ‘a’ called as?....
Question 6. ....
Question 7. ....
Question 8. ....
Question 9. ....
Question 10. ....
Submit Solution