If the integral ∫ ( 8 x+1)/( 2 x- 2 x) d x=A 8 x+k where k is an arbitrary constant, then A is equal to:

Correct answer: A. -(1)/(16) Let I=∫ ( 8 x+1)/( 2 x- 2 x) d x Now, D^r= 2 x- 2 x=( 2 x)/( 2 x)-( 2 x)/( 2 x)=( ^2 2 x- ^2 2 x)/( 2 x 2 x)=(2 4 x)/( 4 x) aligned I &=∫ (2 ^2 4 x)/((2 4 x)/( 4 x)) d x=∫ (2 ^2 4 x · 4 x)/(2 4 x) d x &=(1)/(2) ∫ 8 x d x=-(1)/(2) ( 8 x)/(8)+k &=-(1)/(16) · 8 x+k & Now ,-(1)/(16) · 8 x+k=A 8 x+k & A=-(1)/(16) aligned

JEE Main 201325 AprilOnline

Exam: JEE Main
Year: 2013
Date: 25 April
Mode: Online
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2013 · 25 April · Online · Mathematics · Indefinite Integration

If the integral ∫ ( 8 x+1)/( 2 x- 2 x) d x=A 8 x+k where k is an...

If the integral

\int \frac{\cos 8 x+1}{\cot 2 x-\tan 2 x}

d x=A \cos 8 x+k where

k

is an arbitrary constant, then

\mathrm{A}

is equal to:

Options

A $-\frac{1}{16}$
B $\frac{1}{16}$
C $\frac{1}{8}$
D $-\frac{1}{8}$

Correct answer

A. $-\frac{1}{16}$

Step-by-step solution

Let

\mathrm{I}=\int \frac{\cos 8 x+1}{\cot 2 x-\tan 2 x} d x

Now,

\mathrm{D}^r=\cot 2 x-\tan 2 x=\frac{\cos 2 x}{\sin 2 x}-\frac{\sin 2 x}{\cos 2 x}=\frac{\cos ^2 2 x-\sin ^2 2 x}{\sin 2 x \cos 2 x}=\frac{2 \cos 4 x}{\sin 4 x}

\begin{aligned} \therefore \mathrm{I} &=\int \frac{2 \cos ^2 4 x}{\frac{2 \cos 4 x}{\sin 4 x}} d x=\int \frac{2 \cos ^2 4 x \cdot \sin 4 x}{2 \cos 4 x} d x \\ &=\frac{1}{2} \int \sin 8 x d x=-\frac{1}{2} \frac{\cos 8 x}{8}+k \\ &=-\frac{1}{16} \cdot \cos 8 x+k \\ & \text{ Now },-\frac{1}{16} \cdot \cos 8 x+k=\mathrm{A} \cos 8 x+k \\ & \Rightarrow \mathrm{A}=-\frac{1}{16} \end{aligned}

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