SATHEE: Indefinite Integration 1 Question 86

Indefinite Integration 1 Question 86

87. (i) Show that $\int_{0}^{π} x f (\sin x) d x = \frac{π}{2} \int_{0}^{π} f (\sin x) d x$ .

(ii) Find the value of $\int_{- 1}^{3 / 2} | x \sin π x | d x$ .

$(1982, 2 M)$

$(1982, 3 M)$

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Answer:

Correct Answer: 87. (ii) $\frac{3 π + 1}{π^{2}}$

Solution:

(i) Let $I = \int_{0}^{π} x f (\sin x) d x$

$\Rightarrow I = \int_{0}^{π} (π - x) f (\sin x) d x$

On adding Eqs. (i) and (ii), we get

$2 I = \int_{0}^{π} π f (\sin x) d x$

$∴ \int_{0}^{π} x f (\sin x) d x = \frac{π}{2} \int_{0}^{π} f (\sin x) d x$

(ii) Let $I = \int_{- 1}^{3 / 2} | x \sin π x | d x$

Since, $| x \sin π x | = \begin{array}{cc} x \sin π x, & - 1 < x \leq 1 - x \sin π x, & 1 < x < \frac{3}{2} \end{array}$

$∴ I = \int_{- 1}^{1} x \sin π x d x + \int_{1}^{3 / 2} - x \sin π x d x$

$= 2 - \frac{x \cos π x}{π}_{0}^{1} - 2 \int_{0}^{1} 1 \cdot \frac{- \cos π x}{π} d x$

$- \frac{- x \cos π x}{π}_{1}^{3 / 2} - \int_{1}^{3 / 2} \frac{- \cos π x}{π} d x$

$= \frac{2}{π} + \frac{2}{π^{2}} (0 - 0) + \frac{1}{π} + \frac{1}{π^{2}} (+ 1 - 0)$

$= \frac{3}{π} + \frac{1}{π^{2}} = \frac{3 π + 1}{π^{2}}$

Indefinite Integration 1 Question 86

87. (i) Show that $\int_{0}^{π} x f (\sin x) d x = \frac{π}{2} \int_{0}^{π} f (\sin x) d x$ .

Answer:

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Indefinite Integration 1 Question 86

87. (i) Show that ∫0πxf(sin⁡x)dx=π2∫0πf(sin⁡x)dx.

Answer:

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87. (i) Show that $\int_{0}^{π} x f (\sin x) d x = \frac{π}{2} \int_{0}^{π} f (\sin x) d x$ .