UPSC NDA Math Model Paper 5

Here we have to find the value of ∫

sec2x

√16+tan2x

dx
Let tan‌x=t then sec2xdx=dt
⇒∫

sec2x

√16+tan2x

dx=∫

dt

√16+t2

Now by comparing ∫

dt

√16+t2

with ∫

dx

√x2+a2

we get, a=4
As we know that, ∫

dx

√x2+a2

=log‌x+|√x2+a2|+C where C is a constant
⇒∫

dt

√16+t2

=log|t+√t2+16|+C
Now by substituting tan‌x=t in the above equation we get,
∫

sec2x

√16+tan2x

dx =log|tan‌x+√16+tan2x|+C where C is a constant