TS EAMCET · Maths · Indefinite Integration
- A
- B
- C
- D
Answer & Solution
Correct Answer
(D)
Step-by-step Solution
Detailed explanation
J=∫0π2x3sinxdx=(−x3cosx)0π2+∫0π23x2cosxdx =0+3{(x2sinx)0π2−∫0π22xsinxdx} =3{(π24)−2{(−2cosx)0π2+∫0π2cosx}} =3(π24−2×1)=3π24−6
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