RD Sharma Class 12 Exercise 18.11 Indefinite Integrals Solutions Maths

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RD Sharma Class 12th Exercise 18.11 explains the chapter ‘Indefinite Integrals.’ The exercise contains 12 sums that are of Level 1 difficulty and concept-oriented. However, they are pretty easy to understand how to integrate a function w.r.t to x and also cover trigonometric integrals of values like Tan, Sec, Cosec, Cot. Students should refer to the theorems of those trigonometric values to solve these questions.

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise:18.11
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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.11

Indefinite Integrals Exercise 18.11 Question 2

Answer: $\frac{1}{2} \tan ^{2} x+\frac{1}{4} \tan ^{4} x+C$
Hint :- Use substitution method to solve this integral.
Given :- $\int \tan x \cdot \sec ^{4} x d x$
Sol : - Let $I=\int \tan x \sec ^{4} x d x$
Re-Write $\mathrm{I}=\int \tan x \sec ^{2} x \cdot \sec ^{2} x d x$
$\mathrm{I}=\int \tan x\left(1+\tan ^{2} x\right) \sec ^{2} x d x$ $\text { (if, } \left.\sec ^{2} x=1+\tan ^{2} x\right)$
$\mathrm{I}=\int\left(\tan x+\tan ^{3} x\right) \sec ^{2} x d x$
Substitute $\tan x=t \rightarrow \sec ^{2} x d x=d t$ then
$\mathrm{I}=\int\left(\mathrm{t}+t^{3}\right) \sec ^{2} x \frac{d t}{\sec ^{2} x} \quad \text { (if, } \tan \mathrm{x}=\mathrm{t} \text { ) }$
$=\int\left(\mathrm{t}+t^{3}\right) d t=\int \mathrm{t} d t+\int t^{3} d t$
$=\frac{t^{1+1}}{1+1}+\frac{t^{3+1}}{3+1}+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+\mathrm{C}\right)$ (i)
$I=\frac{t^{2}}{2}+\frac{t^{4}}{4}+C$
$I=\frac{\tan ^{2} x}{2}+\frac{\tan ^{4} x}{4}+\mathrm{C}$ $\text { (if, } t=\tan x)$

Indefinite Integrals Exercise 18.11 Question 3

Answer: $\frac{1}{6} \tan ^{6} x+\frac{1}{8} \tan ^{8} x+C$
Hint: Use substitution method to solve this integral.
Given: $\int \tan ^{5} x \cdot \sec ^{4} x d x$
Solution: Let, $\mathrm{I}=\int \tan ^{5} x \sec ^{4} x d x$
Re-Write $I=\int \tan ^{5} x \sec ^{2} \cdot \sec ^{2} x d x$
$\left.I=\int \tan ^{5} x\left(1+\tan ^{2} x\right) \cdot \sec ^{2} x d x \quad \text { (if, } \sec ^{2} x=1+\tan ^{2} x\right)$
$\mathrm{I}=\int\left(\tan ^{5} x+\tan ^{7} x\right) \sec ^{2} x d x$
$\text { Substitute } \tan \mathrm{x}=\mathrm{t} \rightarrow \sec ^{2} \mathrm{x} \mathrm{d} \mathrm{x}=\mathrm{dt} \text { , then }$
$\left.I=\int\left(\mathrm{t}^{5}+t^{7}\right) \sec ^{2} x \frac{d t}{\sec ^{2} x} \quad \text { (if, } \tan \mathrm{x}=\mathrm{t}\right)$
$=\int\left(\mathrm{t}^{5}+t^{7}\right) d t=\int \mathrm{t}^{5} d t+\int t^{7} d t$
$=\frac{t^{5+1}}{5+1}+\frac{t^{7+1}}{7+1}+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right)$ ......(i)
$\mathrm{I}=\frac{t^{6}}{6}+\frac{t^{8}}{8}+\mathrm{C}$
$\left.\mathrm{I}=\frac{\tan ^{6} x}{6}+\frac{\tan ^{8} x}{8}+\mathrm{C}\: \: \: \: \: \: \: \: \: \quad \text { (if, } \mathrm{t}=\tan \mathrm{x}\right)$

Indefinite Integrals Exercise 18.11 Question 4

Answer: $\frac{1}{6} \sec ^{6} x+C$
Hint:- Use substitution method to solve this integral.
Given: $\int \sec ^{6} x \cdot \tan x d x$
Solution:Let, $\mathrm{I}=\int \sec ^{6} x \cdot \tan x d x$
Re-write, $I=\int \sec ^{5} x \sec x \cdot \tan x d x$ $\mathrm{I}=\int \sec ^{5} x(\sec x \cdot \tan x) d x$

Substitute, $\sec x=t \Rightarrow \sec x \cdot \tan x d x=d t$ , then,

$I=\int t^{5} \cdot \mathrm{dt} \quad(\text { if }, \sec x=t)$

$=\frac{t^{5+1}}{5+1}+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right)$

$\left.=\frac{t^{6}}{6}+c=\frac{\sec ^{6} x}{6}+C \quad \text { (if, } \sec x=t\right)$

Indefinite Integrals Exercise 18.11 Question 5

Answer: $\frac{\tan ^{4} x}{4}-\frac{\tan ^{2} x}{2}+\log |\sec \times| C$
Hint:- Use substitution method to solve this integral.
Given: $\int \tan ^{5} x d x$
Solution: Let, $\mathrm{I}=\int \tan ^{5} x d x$
Re-Write, $\mathrm{I}=\int \tan ^{3} x \cdot \tan ^{2} x \cdot d x$
$I=\int \tan ^{3} x\left(\sec ^{2} x-1\right) d x$ $\text { (if, } \left.\tan ^{2} x=\sec ^{2} x-1\right)$
$\mathrm{I}=\int\left(\tan ^{3} x \cdot \sec ^{2} x-\tan ^{3}\right) d x$
$I=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int \tan ^{3} x d x$
$=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int \tan ^{2} x \cdot \tan x d x$
$\left.=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int\left(\sec ^{2} x-1\right) \cdot \tan x d x \quad \text { (if, } \sec ^{2} x-\tan ^{2} x=1\right)$
$=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int\left(\sec ^{2} x \tan x-\tan x\right) d x$
$=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int \sec ^{2} x \tan x d x+\int \tan x d x$
Put $tan x = t \Rightarrow \sec^{2}x d x = d t , then$
$\mathrm{I}=\int \mathrm{t}^{3} d t-\int t d t+\int \tan x d x$
$=\frac{t^{8+1}}{3+1}+\frac{t^{1+1}}{1+1}+\log |\sec \times| C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+C \& \int \tan x d x=\log |\sec x|+c\right)$
$=\frac{t^{4}}{4}-\frac{t^{2}}{2}+\log |\sec x| C$
$\left.=\frac{\tan ^{4} x}{4}-\frac{\tan ^{2} x}{2}+\log |\sec x| C\: \: \: \: \: \: \quad \text { (if, } t=\tan x\right)$

Indefinite Integrals Exercise 18.11 Question 6

Answer: $\frac{2}{3} \tan ^{\frac{3}{2}} x+\frac{2}{7} \tan ^{\frac{7}{2}} x+C$
Hint: Use substitution method to solve this integral.
Given: $\int \sqrt{\tan x} \cdot \sec ^{4} x d x$
Solution:Let, $I=\int \sqrt{\tan x} \cdot \sec ^{4} x d x$
Re-Write $\mathrm{I}=\int \sqrt{\tan x} \cdot \sec ^{2} x \cdot \sec ^{2} x d x$
$I=\int \sqrt{\tan x}\left(1+\tan ^{2} x\right) \cdot \sec ^{2} x d x \quad\left(\because \sec ^{2} x-\tan ^{2}=1\right)$
$I=\int\left(\sqrt{\tan x}+\sqrt{\tan x} \cdot \tan ^{2} x\right) \sec ^{2} x d x$
$I=\int\left(\tan ^{\frac{1}{2}} x+\tan ^{\frac{1}{2}+2} x\right) \cdot \sec ^{2} x d x$
$I=\int\left(\tan ^{\frac{1}{2}} x+\tan ^{\frac{5}{2}} x\right) \cdot \sec ^{2} x d x$
$\text { Substitute } \tan \mathrm{x}=\mathrm{t} \rightarrow \sec ^{2} \mathrm{x} \mathrm{d} \mathrm{x}=\mathrm{d} \mathrm{t} \text { , then }$
$\begin{array}{ll} \mathrm{I}=\int\left(t^{\frac{1}{2}}+t^{5 / 2}\right) d t \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: & (\because \tan \mathrm{x}=\mathrm{t}) \end{array}$
$=\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{t^{\frac{5}{2}}+1}{\frac{5}{2}+1}+C \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad\left(\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right)$
$=\frac{\tan ^{3 / 2} x}{3 / 2}+\frac{\tan ^{7 / 2} x}{7 / 2}+\mathrm{C} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad(\because \mathrm{t}=\tan \mathrm{x})$
$=\frac{2}{3} \tan ^{3 / 2} x+\frac{2}{7} \tan ^{7 / 2} x+C$

Indefinite Integrals Exercise 18.11 Question 7

$\begin{aligned} &\text { Answer }:-\frac{1}{2} \tan 2 x+\frac{1}{6} \tan ^{3} 2 x+C\\ &\text { Hint :- Use substitution method to solve this integral. }\\ &\text { Given :- } \int \sec ^{4} 2 x d x\\ &\text { Sol: - Let } I=\int \sec ^{4} 2 x d x\\ &\text { Re-writing, } I=\int \sec ^{2} 2 x \cdot \sec ^{2} 2 x d x \end{aligned}$
$\begin{aligned} &\left.I=\int\left(1+\tan ^{2} 2 x\right) \sec ^{2} 2 x d x\: \: \: \: \: \: \: \: \: \: \: \quad \text { (if, } \sec ^{2} x=1+\tan ^{2} x\right) \\ &I=\int\left(\sec ^{2} 2 x+\tan ^{2} 2 x \cdot \sec ^{2} 2 x\right) d x \end{aligned}$
$\begin{aligned} &\text { Substitute } \tan 2 \mathrm{x}=\mathrm{t}\\ &\sec ^{2} 2 x .2 d x=d t, \text { then } \end{aligned}$
$\begin{aligned} &\mathrm{I}=\int \sec ^{2} 2 x \mathrm{~d} x+\int \tan ^{2} 2 x \cdot \sec ^{2} 2 x \mathrm{dx} \\ &=\int \sec ^{2} 2 x \mathrm{~d} x+\int t^{2} \cdot \frac{d t}{2} \: \: \: \: \: \: \: \: \: \: \quad(\text { since }, \tan 2 x=t) \end{aligned}$
$=\frac{\tan 2 x}{2}+\frac{1}{2} \frac{t^{2+1}}{2+1}+C$ $\text { (if, } \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \quad \& \int \sec ^{2} d x=\tan x+\mathrm{c}$
$\begin{aligned} &=\frac{\tan 2 x}{2}+\frac{1}{2} \cdot \frac{t^{3}}{3}+C \\ &\left.=\frac{\tan 2 x}{2}+\frac{1}{6} \cdot \tan ^{3} 2 x+C \: \: \: \: \: \: \: \: \: \: \quad \text { (if, } t=\tan 2 x\right) \end{aligned}$

Indefinite Integrals Exercise 18.11 Question 8

Answer: $-\frac{1}{9} \cot ^{3} 3 x-\frac{1}{3} \cot 3 x+C$
Hint :- Use substitution method to solve this integral.
Given: $\int \operatorname{cosec}^{4} 3 x d x$
Solution:Let $\mathrm{I}=\int \operatorname{cosec}^{4} 3 x d x$
Re-Write $\mathrm{I}=\int \operatorname{cosec}^{2} 3 x \cdot \operatorname{cosec}^{2} 3 x d x$
$\left.I=\int\left(1+\cot ^{2} 3 x\right) \cdot \operatorname{cosec}^{2} 3 x d x \: \: \: \: \: \: \: \: \quad \text { (if, } \operatorname{cosec}^{2} x-\cot ^{2} x=1\right)$
$\begin{aligned} &I=\int\left(\operatorname{cosec}^{2} 3 x+\cot ^{2} 3 x \cdot \operatorname{cosec}^{2} 3 x\right) d x \\ &I=\int \operatorname{cosec}^{2} 3 x \cdot \cot ^{2} 3 x d x+\int \operatorname{cosec}^{2} 3 x d x \end{aligned}$
Substitute, cot3x = t
$\begin{aligned} &\operatorname{cosec}^{2} 3 x \cdot 3 d x=d t, \text { then } \\ &I=\int t^{2} \frac{d t}{3}+\int \operatorname{cosec}^{2} 3 x d x \\ &=-\frac{1}{3} \int t^{2} d t+\int \operatorname{cosec}^{2} 3 x d x \end{aligned}$
$=-\frac{1}{3} \frac{t^{2+1}}{2+1}+\frac{-\cot 3 x}{3}+C$ $\text { (if, } \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \quad \& \int \operatorname{cosec}^{2} x d x=-\cot x+c$
$=-\frac{1}{3} \cdot \frac{t^{3}}{3}-\frac{\cot 3 x}{3}+C$
$=-\frac{1}{9} \cot ^{3} 3 x-\frac{1}{3} \cdot \cot 3 x+\mathrm{C} \: \: \: \: \: \: \: \: \quad(\text { if, } \cot 3 \mathrm{x}=\mathrm{t})$

Indefinite Integrals Exercise 18.11 Question 9

Answer: $-\frac{\cot ^{n+1} x}{n+1}+\mathrm{C}$
Hint:Use substitution method to solve this integral.
Given: $\int \cot ^{n} x \cdot \operatorname{cocec}^{2} x d x, \mathrm{n} \neq-1$
Solution:let, $\mathrm{I}=\int \cot ^{n} x \cdot \operatorname{cocec}^{2} x d x$
Substitute, cot x = t
$\operatorname{cosec}^{2} x d x=\text { dt then }$
$\begin{array}{ll} \mathrm{I}=\int t^{n} \cdot \operatorname{cosec}^{2} \mathrm{x} \cdot \frac{d t}{-\operatorname{cosec}^{2} x} &\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \text { (if, } \cot \mathrm{x}=\mathrm{t}) \end{array}$
$=-\int t^{n}$
$=-\left[\frac{t^{n+1}}{n+1}\right]+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right)$
$\left.=-\frac{\cot ^{n+1} x}{n+1}+C \: \: \: \: \: \: \: \quad \text { (if, } \mathrm{t}=\cot \mathrm{x}\right)$

Indefinite Integrals Exercise 18.11 Question 10

Answer: $-\frac{1}{6} \cot ^{6} x-\frac{1}{8} \cot ^{8} x+C$
Hint :- Use substitution method to solve this integral.
Given: $\int \cot ^{5} x \cdot \operatorname{cosec}^{4} x d x$
Solution:Let, $\mathrm{I}=\int \cot ^{5} x \cdot \operatorname{cosec}^{4} x d x$
Re-Write, $\mathrm{I}=\int \cot ^{5} x \cdot \operatorname{cosec}^{2} x \cdot \operatorname{cosec}^{2} x d x$
$\mathrm{I}=\int \cot ^{5} x\left(1+\cot ^{2} x\right) \cdot \operatorname{cosec}^{2} x d x$ $\text { (if, } \left.\operatorname{cosec}^{2} x-\cot ^{2} x=1\right)$
$\mathrm{I}=\int\left(\cot ^{5} x+\cot ^{5} x \cdot \cot ^{2}\right) \operatorname{cosec}^{2} x d x$
$\mathrm{I}=\int\left(\cot ^{5} x+\cot ^{7} x\right) \operatorname{cosec}^{2} x d x$

Substitute, cotx = t

$-\operatorname{cosec}^{2} x d x=d t, \text { then }$

$I=\int\left(t^{5}+t^{7}\right) \operatorname{cosec}^{2} x \cdot \frac{d t}{-\operatorname{cosec}^{2} x} \mathrm{dx} \: \: \: \: \: \: \: \: \quad(\text { as } \cot x=t)$

$\begin{aligned} &=-\int\left(t^{5}+t^{7}\right) \mathrm{dt} \\ &=-\int t^{5} d t-\int t^{7} \mathrm{dt} \end{aligned}$

$=-\frac{t^{5+1}}{5+1}-\frac{t^{7+1}}{7+1}+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+\mathrm{c}\right)$

$=-\frac{t^{6}}{6}-\frac{t^{8}}{8}+C$

$=-\frac{\cot ^{6} x}{6}-\frac{\cot ^{8} x}{8}+\mathrm{C}$ $(\text { if, } t=\cot x)$

Indefinite Integrals Exercise 18.11 Question 11

Answer: $-\frac{1}{4} \cot ^{4} x+\frac{1}{2} \cot ^{2} x+\log |\sin x|+C$
Hint :- Use substitution method to solve this integral.
Given: $\int \cot ^{5} x d x$
Solution: Let $I=\int \cot ^{5} x d x$
Re-Write, $\mathrm{I}=\int \cot ^{5} x d x$
$=\int \cot ^{3} x \cdot \cot ^{2} x d x$
$\mathrm{I}=\int \cot ^{3} x\left(\operatorname{cosec}^{2} x-1\right) d x$ $\text { (if, } \operatorname{cosec}^{2} x-\cot ^{2} x=1 \text { ) }$
$\begin{aligned} &I=\int\left(\cot ^{3} x \cdot \operatorname{cosec}^{2} x-\cot ^{3} x\right) d x \\ &I=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x d x-\int \cot ^{3} x d x \\ &I=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x d x-\int \cot ^{2} x \cot x d x \end{aligned}$
$\begin{aligned} &\mathrm{I}=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x d x-\int\left(\operatorname{cosec}^{2} x-1\right) \cot x d x \\ &\mathrm{I}=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x d x-\int\left(\operatorname{cosec}^{2} x \cdot \cot x-\cot x\right) d x \\ &\mathrm{I}=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x d x-\int \operatorname{cosec}^{2} x \cdot \cot x d x+\int \cot x d x \end{aligned}$

Substitute, cotx = t

$\begin{aligned} &-\operatorname{cosec}^{2} x d x=d t, \text { then } \\ &I=\int t^{3}(-d t)-\int t(-d t)+\int \cot x d x \\ &=-\int t^{3} d t+\int t d t+\int \cot x d x \end{aligned}$

$=-\frac{t^{8+1}}{3+1}+\frac{t^{1+1}}{1+1}+\log |\sin x|+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+\mathrm{c} \& \int \cot x d x=\log |\sin x|+c\right)$

$\begin{aligned} &=-\frac{1}{4} t^{4}+\frac{1}{2} t^{2}+\log |\sin x|+C \\ &=-\frac{1}{4} \cot ^{4} x+\frac{1}{2} \cot ^{2} x+\log |\sin x|+C \end{aligned}$ $\left ( as,t=\cot x \right )$

Indefinite Integrals Exercise 18.11 Question 12

Answer: $-\frac{1}{5} \cot ^{5} x+\frac{1}{3} \cot ^{3} x-\cot x-x+C$
Hint :- Use substitution method to solve this integral.
Given: $\int \cot ^{6} x d x$
Solution: Let, $I=\int \cot ^{6} x d x$
Re-Write, $\mathrm{I}=\int \cot ^{6} x d x$
$\mathrm{I}=\int \cot ^{2} x \cdot \cot ^{4} x d x$
$\mathrm{I}=\int\left(\operatorname{cosec}^{2} x-1\right) \cot ^{4} x d x$ $\text { (if, } \operatorname{cosec}^{2} x-\cot ^{2} x=1 \text { ) }$
$\begin{aligned} &I=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{4} x d x \\ &I=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x \cdot \cot ^{2} x d x \end{aligned}$
$\left.I=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x\left(\operatorname{cosec}^{2} x-1\right) d x \quad \text { (if, } \operatorname{cosec}^{2} x-\cot ^{2} x=1\right)$
$\begin{aligned} &1=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int\left(\cot ^{2} x \cdot \operatorname{cosec}^{2} x-\cot ^{2} x\right) d x \\ &\mathrm{I}=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x \cdot \operatorname{cosec}^{2} x d x+\int \cot ^{2} x d x \end{aligned}$
$\left.I=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x \cdot \operatorname{cosec}^{2} x d x+\int\left(\operatorname{cosec}^{2} x-1\right) d x \text { (if, } \cot ^{2} x=\operatorname{cosec}^{2} x-1\right)$
$\mathrm{I}=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x \cdot \operatorname{cosec}^{2} x d x+\int \operatorname{cosec}^{2} x d x-\int 1 d x$
Substitute, cotx = t
$-\operatorname{cosec}^{2} x d x=d t, \text { then }$
$\begin{aligned} &\mathrm{I}=\int t^{4}(-d t)-\int t^{2}(-d t)+\int \operatorname{cosec}^{2} x d \mathrm{x}-\int 1 \mathrm{dx} \\ &=-\int t^{4} d t+\int t^{2} d t+\int \operatorname{cosec}^{2} x d \mathrm{x}-\int x^{0} \mathrm{dx} \end{aligned}$
$=-\frac{t^{4+1}}{4+1}+\frac{t^{2+1}}{2+1}-\cot x-\frac{x^{0+1}}{0+1}+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+c \& \int \operatorname{cosec}^{2} x d x=-\cot x+c\right)$
$\begin{aligned} &=-\frac{t^{5}}{5}+\frac{t^{3}}{3}-\cot x-x+C \\ &=-\frac{\cot ^{5} x}{5}+\frac{\cot ^{8} x}{3}-\cot x-x+C \end{aligned}$ $\left ( if,\cot x=t \right )$

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