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AP EAMCET · Maths · Indefinite Integration

If In=0π4tannxdx, then 1I2+14+1I3+I5+1I4+16=

  1. A 1I9+I11
  2. B 1I10+I12
  3. C 1I12+I14
  4. D 1I11+I13
Verified Solution

Answer & Solution

Correct Answer

(D) 1I11+I13

Step-by-step Solution

Detailed explanation

Given In=∫0π4tannxdx In+In+2=∫0π4tannxdx+∫0π4tann+2xdx =∫0π4tannxsec2xdx =∫01tndt where tanx=t =tn+1n+101=1n+1 Now, 1I2+14+1I3+I5+1I4+16=3+4+5=12 =1I11+I13
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