Chapter 4: Q23MP (page 239)

Find $\frac{d}{d t} \int_{0}^{s i n t} \frac{s i n^{- 1} x}{x} d x$ .

Short Answer

Expert verified

The value of $\frac{d}{d t} \int_{0}^{\sin t} \frac{\sin^{- 1} x}{x} d x is t \cot t$ .

Step by step solution

Given Information

Given that . $\frac{d}{d t} \int_{0}^{\sin t} \frac{\sin^{- 1} x}{x} d x$ ...(1)

Formula Used

We know that $\frac{d}{d x} \int_{u (x)}^{v (x)} f (x, t) d t = f (x, v) \frac{d v}{d x} - f (x, u) \frac{d u}{d x} + \int_{u}^{v} \frac{\partial f}{\partial x} d t \in$ ...(2)

Solve the given equation

Compare equation (1) and equation (2).

$\frac{d}{d t} \int_{0}^{\sin t} \frac{\sin^{- 1} x}{x} d x = \frac{\sin^{- 1} (\sin t)}{\sin t} . \frac{d}{d t} (\sin t) - \frac{\sin^{- 1} (0)}{0} . \frac{d}{d t} (0) + \int_{0}^{\sin t} 0 d x$

Therefore,

$\frac{d}{d t} \int_{0}^{\sin t} \frac{\sin^{- 1} x}{x} d x = \frac{t}{\sin t} . (\cos t) \frac{d}{d t} \int_{0}^{\sin t} \frac{\sin^{- 1} x}{x} d x = t (\frac{\cos t}{\sin t}) \frac{d}{d t} \int_{0}^{\sin t} \frac{\sin^{- 1} x}{x} d x = t \cot t$

Hence the value of $\frac{d}{d t} \int_{0}^{\sin t} \frac{\sin^{- 1} x}{x} d x = t \cot t$ .

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Find $\frac{d}{d t} \int_{0}^{s i n t} \frac{s i n^{- 1} x}{x} d x$ .

Short Answer

Step by step solution

Given Information

Formula Used

Solve the given equation

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Short Answer

Step by step solution

Given Information

Formula Used

Solve the given equation

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Most popular questions from this chapter

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Find $\frac{d}{d t} \int_{0}^{s i n t} \frac{s i n^{- 1} x}{x} d x$ .