SATHEE: Indefinite Integration 1 Question 78

Indefinite Integration 1 Question 78

79. Prove that for any positive integer $k$ ,

$\frac{\sin 2 k x}{\sin x} = 2 [\cos x + \cos 3 x + \dots + \cos (2 k - 1) x]$

Hence, prove that $\int_{0}^{π / 2} \sin 2 k x \cdot \cot x d x = π / 2$ .

$(1990, 4$ M)

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Solution:

We know that,

$2 \sin x [\cos x + \cos 3 x + \cos 5 x + \dots + \cos (2 k - 1) x]$

$= 2 \sin x \cos x + 2 \sin x \cos 3 x + 2 \sin x \cos 5 x$

$+ \dots + 2 \sin x \cos (2 k - 1) x$

$= \sin 2 x + (\sin 4 x - \sin 2 x) + (\sin 6 x - \sin 4 x)$

$+ \dots + \sin 2 k x - \sin (2 k - 2) x$

$= \sin 2 k x$

$∴ 2 [\cos x + \cos 3 x + \cos 5 x + \dots + \cos (2 k - 1) x]$

$= \frac{\sin 2 k x}{\sin x}$

Now, $\sin 2 k x \cdot \cot x = \frac{\sin 2 k x}{\sin x} \cdot \cos x$

$= 2 \cos x [\cos x + \cos 3 x + \cos 5 x + \dots + \cos (2 k - 1) x]$

[from Eq. (i)]

$= [2 \cos^{2} x + 2 \cos x \cos 3 x + 2 \cos x \cos 5 x +$ . $+ 2 \cos x \cos (2 k - 1) x]$

$= (1 + \cos 2 x) + (\cos 4 x + \cos 2 x)$

$+ (\cos 6 x + \cos 4 x) + \dots + \cos 2 k x + \cos (2 k - 2) x$

$= 1 + 2 [\cos 2 x + \cos 4 x + \cos 6 x + \dots + \cos (2 k - 2) x]$

$+ \cos 2 k x$

$∴ \int_{0}^{π / 2} (\sin 2 k x) \cdot \cot x d x$

$= \int_{0}^{π / 2} 1 \cdot d x + 2 \int_{0}^{π / 2} (\cos 2 x + \cos 4 x \dots \cos (2 k - 2) x) d x$ $+ \int_{0}^{π / 2} \cos (2 k) x d x$

$= \frac{π}{2} + 2 \frac{\sin 2 x}{2} + \frac{\sin 4 x}{4} + \dots + {\frac{\sin (2 k - 2) x}{(2 k - 2)}}_{0}^{π / 2}$

$+ \frac{\sin (2 k) x}{2 k}_{0}^{π / 2} = \frac{π}{2}$

Indefinite Integration 1 Question 78

79. Prove that for any positive integer $k$ ,

Solution:

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Indefinite Integration 1 Question 78

79. Prove that for any positive integer k,

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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79. Prove that for any positive integer $k$ ,