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Question: How do you evaluate the integral of \(\int {t\sin 2tdt} \)?...

How do you evaluate the integral of ∫tsin⁡2tdt\int {t\sin 2tdt} ?

Explanation

Solution

First of all, we will consider t as the first function and sin (2t) as the second function. Then just proceed with the ILATE rule to integrate the required integral which the help of integration of u.v

Complete step-by-step solution:
We are given that we are required to integrate the t. sin (2t)
Now, we need to find the value of ∫tsin⁡2tdt\int {t\sin 2tdt} .
Now using the ILATE rule, we have t as the first function and sin (2t) as the second function.
Now, we get:-
⇒∫tsin⁡2tdt=−tcos⁡2t2+∫cos⁡2t2dt+C\Rightarrow \int {t\sin 2tdt} = - t\dfrac{{\cos 2t}}{2} + \int {\dfrac{{\cos 2t}}{2}dt + C}
We can write this as follows:-
⇒∫tsin⁡2tdt=−tcos⁡2t2+12∫cos⁡2tdt+C\Rightarrow \int {t\sin 2tdt} = - \dfrac{{t\cos 2t}}{2} + \dfrac{1}{2}\int {\cos 2tdt + C}
Now integrating on the right hand side in above equation, we will then obtain the following equation:-
⇒∫tsin⁡2tdt=−tcos⁡2t2+sin⁡2t2×2+C\Rightarrow \int {t\sin 2tdt} = - \dfrac{{t\cos 2t}}{2} + \dfrac{{\sin 2t}}{{2 \times 2}} + C
Simplifying the right hand side of the above equation further, we will then obtain the following equation with us:-
⇒∫tsin⁡2tdt=−tcos⁡2t2+sin⁡2t4+C\Rightarrow \int {t\sin 2tdt} = - \dfrac{{t\cos 2t}}{2} + \dfrac{{\sin 2t}}{4} + C
Thus, we have the required answer.

Note: The students must know the ILATE rule which has been mentioned above.
I stands for Inverse, L stands for Logarithmic, A stands for Algebraic, T stands for Trigonometric and E stands for exponential. This is the sequence in which we take the first function. Therefore, we have taken t as the first function and sin (2t) as the second function.
Also, note that if there are two function; first function being f (x) and the second function being g (x).
The integration of f (x) and g (x) is given by the following expression:-
\Rightarrow \int {f(x).g(x)dx = f(x)\int {g(x)dx - \int {\left\\{ {\left( {\dfrac{d}{{dx}}f(x)} \right)\int {g(x)dx} } \right\\}dx + C} } }
This is the formula that we used above.
Here, we just took f (x) to be equal to t and g (x) to be equal to sin (2t).
The students must also note that we have used the following facts in the solution mentioned above:-
1. ∫cos⁡nxdx=sin⁡nxn+c\int {\cos nxdx = \dfrac{{\sin nx}}{n} + c}
2. ∫sin⁡nxdx=−cos⁡nxn+c\int {\sin nxdx = \dfrac{{ - \cos nx}}{n} + c}
Here, we just replaced x by t and n by 2 in the above formulas to get the desired results in the solution mentioned above.