SATHEE: Indefinite Integration 1 Question 19

Indefinite Integration 1 Question 19

20. The integral $\int_{π / 4}^{π / 2} (2 cosec x)^{17} d x$ is equal to

(a) $\int_{0}^{\log (1 + \sqrt{2})} 2 {(e^{u} + e^{- u})}^{16} d u$

(2014 Adv.)

(b) $\int_{0}^{\log (1 + \sqrt{2})} {(e^{u} + e^{- u})}^{17} d u$

(d) $\int_{0}^{\log (1 + \sqrt{2})} 2 {(e^{u} - e^{- u})}^{16} d u$

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Solution:

PLAN This type of question can be done using appropriate substitution.

Given, $I = \int_{π / 4}^{π / 2} (2 cosec x)^{17} d x$

$= \int_{π / 4}^{π / 2} \frac{2^{17} (cosec x)^{16} cosec x (cosec x + \cot x)}{(cosec x + \cot x)} d x$

Let $cosec x + \cot x = t$

$\Rightarrow (- cosec x \cdot \cot x - {cosec}^{2} x) d x = d t$

and $cosec x - \cot x = 1 / t$

$\begin{aligned} \Rightarrow & 2 cosec x & = t + \frac{1}{t} \\ ∴ & I & = - \int_{\sqrt{2} + 1}^{1} 2^{17} \frac{t + \frac{1}{t}}{2} \frac{d t}{t} \end{aligned}$

Let $t = e^{u} \Rightarrow d t = e^{u} d u$ .

When $t = 1, e^{u} = 1 \Rightarrow u = 0$

and when $t = \sqrt{2} + 1, e^{u} = \sqrt{2} + 1$

$\begin{aligned} \Rightarrow u & = \ln (\sqrt{2} + 1) \\ \Rightarrow I & = - \int_{\ln (\sqrt{2} + 1)}^{0} 2 {(e^{u} + e^{- u})}^{16} \frac{e^{u} d u}{e^{u}} \\ = 2 \int_{0}^{\ln (\sqrt{2} + 1)} {(e^{u} + e^{- u})}^{16} d u \end{aligned}$

Indefinite Integration 1 Question 19

20. The integral $\int_{π / 4}^{π / 2} (2 cosec x)^{17} d x$ is equal to

Solution:

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Indefinite Integration 1 Question 19

20. The integral ∫π/4π/2(2cosecx)17dx is equal to

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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20. The integral $\int_{π / 4}^{π / 2} (2 cosec x)^{17} d x$ is equal to