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Question: If \[f(x) = \int\limits_0^x {t\sin tdt,} \] then write the value of \[f'(x)\]....

If f(x)=∫0xtsin⁡tdt,f(x) = \int\limits_0^x {t\sin tdt,} then write the value of f′(x)f'(x).

Explanation

Solution

First of all, do the integration using this formula:
∫u⋅vdx=u∫vdx−∫(dudx∫vdx)dx\int {u \cdot vdx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\int v dx} \right)} } } dx
The choice of first function u to be selected will be one, which comes first in the order of L I A T E. Where L, I, A, T and E terms indicate logarithmic, inverse trigonometric, Algebraic, Trigonometric and Exponential function.
Then do the derivative of the f(x) with respect to x and we get the answer.

Complete step by step answer:
The function f(x) given in the expression in form of integration is:
f(x)=∫0xtsin⁡tdt,f(x) = \int\limits_0^x {t\sin tdt,}
Step1:
Now solving this integration by using the below formula of by parts integration.
∫u⋅vdx=u∫vdx−∫(dudx∫vdx)dx\int {u \cdot vdx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\int v dx} \right)} } } dx
Where u=tu = t
v=sin⁡tv = \sin t
=[t∫sin⁡tdt−∫(dtdt∫sin⁡tdt)dt]0x= \left[ {t\int {\sin tdt - \int {\left( {\dfrac{{dt}}{{dt}}\int {\sin tdt} } \right)} } dt} \right]_0^x
=[−tcos⁡t+∫cos⁡tdt]0x= \left[ { - t\cos t + \int {\cos tdt} } \right]_0^x
In the above equation, we have used the relation that the integration of the sin⁡t\sin t with respect to t is −cos⁡t - \cos t.
=[−tcos⁡t+sin⁡t]0x= \left[ { - t\cos t + \sin t} \right]_0^x
In the above equation, we have used the relation that the integration of the cos⁡t\cos t with respect to t is sin⁡t\sin t. Then put the limit in the equation.
=−xcos⁡x+sin⁡x= - x\cos x + \sin x

Step2: Compute the function f′(x)f'(x) by differentiating the f(x)f(x).
f′(x)=ddxf(x)f'(x) = \dfrac{d}{{dx}}f(x)
=ddx(−xcos⁡x+sin⁡x)= \dfrac{d}{{dx}}( - x\cos x + \sin x)
=−(dxdx)(cos⁡x)−(x)(ddxcos⁡x)+ddx(sin⁡x)= - \left( {\dfrac{{dx}}{{dx}}} \right)\left( {\cos x} \right) - (x)\left( {\dfrac{d}{{dx}}\cos x} \right) + \dfrac{d}{{dx}}(\sin x)
In above equation, we have used the relation that the derivation of the sin⁡x\sin x with respect to x is cos⁡x\cos x and the derivation of the cos⁡x\cos x with respect to x is −sin⁡x - \sin x
=−cos⁡x+xsin⁡x+cos⁡x= - \cos x + x\sin x + \cos x
=xsin⁡x= x\sin x
We get the f′(x)=xsin⁡xf'(x) = x\sin x.

Note:
Students are likely to make mistakes while doing integration by parts, students are advised to write the formula and then write the values for u and v and then substitute into the formula to avoid mistakes. Students should make choices of u and v very carefully by LIATE rule and then proceed with the question. LIATE rule is also the same as ILATE rule, it doesn’t matter even if we change the order of the logarithmic function and exponential function.
Alternative method:
Another approach is f′(x)=ddx∫0xf(t)dt=f(x)f'(x) = \dfrac{d}{{dx}}\int\limits_0^x {f(t)dt = } f(x), so from this rule we can directly find the value of f′(x)=xsin⁡xf'(x) = x\sin x by substituting x in place of t and f(x)=f(t)=tsin⁡t=xsin⁡xf(x) = f(t) = t\sin t = x\sin x.