Answer :
Using
sin(π – x) = sinx and cos(π – x) = -cosx
Add (i) and (ii)
Substitute cosx = t
When x = 0 ⇒ t = 1 and when x = π ⇒ t = -1
Differentiate t = cosx with respect to x
⇒ -dt = sinxdx
Put in (a)
Using
⇒ I = 2π[tan-1(1) – tan-1(-1)]
⇒ I = π2
Hence
OR
Multiply and divide by 2,
⇒ I = 1/2(I1 – I2) …(i)
Let us first solve I1
Substitute x2 + 5x + 6 = t
Differentiate with respect to x
⇒ dt = (2x + 5)dx
⇒ I1 = 2√t
Resubstitute t
Now let us find I2
Using the result
Put I1 and I2 value in (i)
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