We can write above integral as

[Adding and subtracting 1 in denominator]

∵ sin²x + cos²x = 1 and

sin2x = 2 sinx cosx

∵sin²x + cos²x + 2 sinx cosx = (sinx + cosx)²

Taking minus (-) common from numerator we get,

Put sinx + cosx = t

Differentiating w.r.t x we get,

(cosx - sinx)dx = dt

Putting values we get,

We know that,

Here x = t and a = 1

Putting value of t we get,

∴ from (1) we get,