UPSEE Solved Model Paper 7

I ∫ sec2θ (sec θ + tan θ) . (sec θ + tan θ) dθ
Put sec θ + tan θ = y ... (1)
∴ sec θ (tan θ + sec θ) dθ = dy
Now, sec θ - tan θ =

1

y

... (2)
From (1) and (2) , 2 sec θ = y+

1

y

∴ I =

1

2

‌∫y(y+

1

y

) dy =

1

2

[

y3

3

+y] + C
=

1

2

[

(secθ+tanθ)3

3

+(secθ+tanθ)] + C
=

(secθ+tanθ)

6

[(secθ+tanθ)2+3] + C