The integral ∫ ^ 2 x ( x+ x)^ (9)/(2) d x equals (for some arbitrary constant K )

Correct answer: C. - 1 ( x+ x)^ (11)/(2) \ (1)/(11)+(1)/(7)( x+ x)^ 2 \ +K array l I=∫ ^ 2 x ( x+ x)^ 9 / 2 d x Let x+ x=t x- x=(1)/(t) x=(1)/(2)(t+(1)/(t)) and x( x+ x) d x=d t x d x=(d t)/(t) I=(1)/(2) ∫ (t+(1)/(t)) d t t^ 9 / 2 · t =(1)/(2) ∫(t^ -9 / 2 +t^ -13 / 2 ) d t =(-1)/(7) t^ -7 / 2 -(1)/(11) t^ -1 / 2 +K =- 1 7 t^ 7 / 2 - 1 11 t^ 11 / 2 +K=- 1 t^ 11 / 2 ((1)/(11)+ t^ 2 7 )+K = -1 ( x+ x)^ 11 / 2 \ (1)/(11)+(1)/(7)( x+ x)^ 2 \ +K array

JEE Advanced 2012Paper 1

Exam: JEE Advanced
Year: 2012
Paper: Paper 1
Subject: Mathematics
Chapter: Indefinite Integration

JEE Advanced 2012 · Paper 1 · Mathematics · Indefinite Integration

The integral ∫ ^ 2 x ( x+ x)^ (9)/(2) d x equals (for some...

The integral

\int \frac{\sec ^{2} x}{(\sec x+\tan x)^{\frac{9}{2}}} d x

equals (for some arbitrary constant

\mathrm{K}

)

Options

A $-\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$
B $\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$
C $-\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$
D $\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$

Correct answer

C. $-\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$

Step-by-step solution

\begin{array}{l} I=\int \frac{\sec ^{2} x}{(\sec x+\tan x)^{9 / 2}} d x \\ \text{ Let } \sec x+\tan x=t \Rightarrow \sec x-\tan x=\frac{1}{t} \\ \Rightarrow \sec x=\frac{1}{2}\left(t+\frac{1}{t}\right) \text{ and } \sec x(\sec x+\tan x) d x=d t \\ \Rightarrow \sec x d x=\frac{d t}{t} \\ \therefore \quad I=\frac{1}{2} \int \frac{\left(t+\frac{1}{t}\right) d t}{t^{9 / 2} \cdot t}=\frac{1}{2} \int\left(t^{-9 / 2}+t^{-13 / 2}\right) d t \\ =\frac{-1}{7} t^{-7 / 2}-\frac{1}{11} t^{-1 / 2}+K \\ =-\frac{1}{7 t^{7 / 2}}-\frac{1}{11 t^{11 / 2}}+K=-\frac{1}{t^{11 / 2}}\left(\frac{1}{11}+\frac{t^{2}}{7}\right)+K \\ =\frac{-1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K \end{array}

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