SATHEE: Indefinite Integration 1 Question 14

Indefinite Integration 1 Question 14

15. The value of $\int_{0}^{π} | \cos x |^{3} d x$ is

(2019 Main, 9 Jan I)

(a) $\frac{2}{3}$

(b) $- \frac{4}{3}$

(d) $\frac{4}{3}$

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Solution:

We know, graph of $y = \cos x$ is

$∴$ The graph of $y = | \cos x |$ is

$I = \int_{0}^{π} | \cos x |^{3} = 2 \int_{0}^{\frac{π}{2}} | \cos x |^{3} d x$

$(∵ y = | \cos x |$ is symmetric about $x = \frac{π}{2})$

$= 2 \int_{0}^{\frac{π}{2}} \cos^{3} x d x ∵ \cos x \geq 0 for x \in 0, \frac{π}{2}$

Now, as $\cos 3 x = 4 \cos^{3} x - 3 \cos x$

$∴ \cos^{3} x = \frac{1}{4} (\cos 3 x + 3 \cos x)$

$∴ I = \frac{2}{4} \int_{0}^{\frac{π}{2}} (\cos 3 x + 3 \cos x) d x$

$= \frac{1}{2} \frac{\sin 3 x}{3} + 3 \sin x_{0}^{\frac{π}{2}}$

$= \frac{1}{2} \frac{1}{3} \sin \frac{3 π}{2} + 3 \sin \frac{π}{2} - \frac{1}{3} \sin 0 + 3 \sin 0$

$= \frac{1}{2} \frac{1}{3} (- 1) + 3 - [0 + 0]$

$∵ \sin \frac{3 π}{2} = \sin π + \frac{π}{2} = - \sin \frac{π}{2} = - 1$

$= \frac{1}{2} - \frac{1}{3} + 3 = \frac{4}{3}$

Indefinite Integration 1 Question 14

15. The value of $\int_{0}^{π} | \cos x |^{3} d x$ is

Solution:

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Indefinite Integration 1 Question 14

15. The value of ∫0π|cos⁡x|3dx is

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

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Other Resources

Syllabus

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Sample Mock Test

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NCERT Exercise Video Solution

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15. The value of $\int_{0}^{π} | \cos x |^{3} d x$ is