JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Indefinite Integrals

The integral

\int \frac{s e c ^{2} x}{( s e c x + t a n x ) ^{9/2}} d x

equals (for some arbitrary constant

K

)

(A) $- \frac{1}{( s e c x + t a n x ) ^{11/2}} {\frac{1}{11} - \frac{1}{7} (sec x + tan x)^{2}} + K$
(B) $\frac{1}{( s e c x + t a n x ) ^{11/2}} {\frac{1}{11} - \frac{1}{7} (sec x + tan x)^{2}} + K$
(C) $- \frac{1}{( s e c x + t a n x ) ^{11/2}} {\frac{1}{11} + \frac{1}{7} (sec x + tan x)^{2}} + K$ ✓ Correct
(D) $\frac{1}{( s e c x + t a n x ) ^{11/2}} {\frac{1}{11} + \frac{1}{7} (sec x + tan x)^{2}} + K$

Answer: (C)

Solution

Let

t = sec x + tan x

. Then

d t = (sec x tan x + sec^{2} x) d x = sec x (sec x + tan x) d x = t sec x d x

, so

sec x d x = \frac{d t}{t}

. Also,

sec x = \frac{1}{2} (t + \frac{1}{t})

. The integral becomes

I = \frac{1}{2} \int (t^{- 9/2} + t^{- 13/2}) d t = \frac{1}{2} [\frac{t ^{- 7/2}}{- 7/2} + \frac{t ^{- 11/2}}{- 11/2}] = - \frac{1}{7 t ^{7/2}} - \frac{1}{11 t ^{11/2}} = - \frac{1}{t ^{11/2}} (\frac{1}{11} + \frac{t ^{2}}{7}) + K

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Single correct option