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TS EAMCET · Maths · Definite Integration

The value of 0π/2dx1+tanx2018 is equal to

  1. A π
  2. B 3π4
  3. C π2
  4. D π4
Verified Solution

Answer & Solution

Correct Answer

(D) π4

Step-by-step Solution

Detailed explanation

Let I=∫0π211+tanx2018dx ⇒I=∫0π211+sinxcosx2018dx ⇒I=∫0π2cosx2018cosx2018+sinx2018dx       …i ⇒I=∫0π2cosπ2-x2018cosπ2-x2018+sinπ2-x2018dx…
From TS EAMCET
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