Chapter 4: Q3P (page 237)

$If z = \int_{sinx}^{^{cosx}} \frac{\sin t}{dt}, find \frac{dz}{dx} .$

Short Answer

Expert verified

The value of $\frac{dz}{dx}$ islocalid="1659235323161" $- \sin (\cos x) \tan x - \sin (\sin x) cotx$ .

Step by step solution

Given Information

Given the value of $z = \int_{sinx}^{^{cosx}} \frac{\sin t}{t} dt$ …. (1)

Finding dzdx

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

On Compare equation (1) and equation (2), then the given function become $\frac{d}{d x} \int_{\sin x}^{\cos x} \frac{\sin t}{t} = \frac{\sin (\cos x)}{c o s x} . \frac{d}{d x} (c o s x) - \frac{\sin (\sin x)}{s i n x} . \frac{d}{d x} (s i n x) + \int_{\sin x}^{\cos x} 0 . d t \frac{d}{d x} \int_{\sin x}^{\cos x} \frac{\sin t}{t} d t = \frac{\sin (\cos x)}{c o s x} . (- s i n x) - \frac{\sin (\sin x)}{s i n x} (c o s x) \frac{d}{d x} \int_{\sin x}^{\cos x} \frac{\sin t}{t} d t = \sin (\cos x) (\frac{- \sin x}{\cos x}) - \sin (\sin x) (\frac{\cos x}{\sin x}$

Therefore

$\frac{d}{dx} \int_{\sin x}^{\cos x} \frac{\sin t}{t} dt = \sin (\cos x) (- \tan x) - \sin (\sin x) (\cot x) \frac{d}{dx} \int_{\sin x}^{\cos x} \frac{\sin t}{t} dt = - \sin (\cos x) \tan x - \sin (\sin x) \cot x$

Therefore, $\frac{dz}{dx} = - \sin (\cos x) \tan x - \sin (\sin x) \cot x .$

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$If z = \int_{sinx}^{^{cosx}} \frac{\sin t}{dt}, find \frac{dz}{dx} .$

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Finding dzdx

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Ifz=∫sinxcosxsintdt,finddzdx.

Short Answer

Step by step solution

Given Information

Finding dzdx

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Most popular questions from this chapter

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$If z = \int_{sinx}^{^{cosx}} \frac{\sin t}{dt}, find \frac{dz}{dx} .$