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What is the value of \(\mathop \smallint \nolimits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} \;x\sin x\;dx\;\)
5. None of these

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Correct Answer - Option 2 : 2

Concept:

I. Integration by parts:

Let f(x) and g(x) be two differentiable functions of x.

Then \(\smallint f\left( x \right)\;g\left( x \right)\;dx = f\left( x \right)\;\left[ {\smallint g\left( x \right)\;dx} \right] - \;\smallint \left[ {\frac{d}{{dx}}\left( {f\left( x \right)} \right) \times \;\smallint g\left( x \right)\;dx} \right]\) where f(x) is the first function and g(x) is the second function.

These functions are ordered on the basis of “ILATE” i.e inverse trigonometric function, logarithmic function, algebraic function, trigonometric function, exponential function.

II. \(\mathop \smallint \nolimits_{ - \;a}^a f\left( x \right)\;dx = \;\left\{ {\begin{array}{*{20}{c}} {2 \times \;\mathop \smallint \nolimits_0^a f\left( x \right),\;\text{if}\;\;f\left( x \right)\;\text{is even function}}\\ {0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{if}\;f\left( x \right)\;\text{is an odd function}} \end{array}} \right.\) 

III. Let f be a function. Then f is said to be an even function if f(- x) = f(x) and odd function if f(- x) = - f(x).

Calculation:

\(\mathop \smallint \nolimits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} \;x\sin x\;dx\;\)

Here, f(x) = x sin x

⇒ f(- x) = (- x) × sin (- x) = x × sin x      ----(∵ sin x is an odd function)

⇒ f(- x) = f(x)

Hence, the given function f(x) is an even function.

As we know that, \(\mathop \int \nolimits_{ - \;a}^a f\left( x \right)\;dx = \;\left\{ {\begin{array}{*{20}{c}} {2 \times \;\mathop \smallint \nolimits_0^a f\left( x \right),\; \text{if}\;\;f\left( x \right)\;\text{is even function}}\\ {0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{if}\;f\left( x \right)\; \text{is an odd function}} \end{array}} \right.\)

\(\Rightarrow \mathop \smallint \nolimits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} \;x\sin x\;dx\; = \;2\mathop \smallint \nolimits_0^{\frac{\pi }{2}} x\sin x\;dx\)

According to the ILATE rule, the first function is x i.e f(x) = x and second function is sin x i.e g(x) = sin x

As we know that, \(\smallint f\left( x \right)\;g\left( x \right)\;dx = f\left( x \right)\;\left[ {\smallint g\left( x \right)\;dx} \right] - \;\smallint \left[ {\frac{d}{{dx}}\left( {f\left( x \right)} \right) \times \;\smallint g\left( x \right)\;dx} \right]\;dx\)

\(\Rightarrow \mathop \smallint \nolimits_0^{\frac{\pi }{2}} x\sin x\;dx = x \times \left[ {\mathop \smallint \nolimits_0^{\frac{\pi }{2}} \sin x\;dx} \right] - \;\mathop \smallint \nolimits_0^{\frac{\pi }{2}} \left[ {\frac{{d\left( x \right)}}{{dx}} \times \;\;\mathop \smallint \nolimits_0^{\frac{\pi }{2}} \sin x\;dx} \right]dx\)

\(\Rightarrow 2\mathop \smallint \nolimits_0^{\frac{\pi }{2}} x\sin x\;dx = 2\;{\left[ {x\;\left( { - \cos x} \right) + \;\smallint \cos x\;dx} \right]_0}^{\frac{\pi }{2}}\)

\(\Rightarrow \;\mathop \smallint \nolimits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} \;x\sin x\;dx\; = 2\)

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