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Q11E

Expert-verifiedFound in: Page 326

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**Evaluate the integral: \(\int {t{{\sin }^2}tdt} \)**

To evaluate the integral \(\int {t{{\sin }^2}tdt} \) , we will use identity: \({\sin ^2}t = \frac{{1 - \cos 2t}}{2}\) followed by the integration formula given by \(\int {uvdx = u\int {vdx - \int {\frac{{du}}{{dx}}\left( {\int {vdx} } \right)dx} } } \)

Firstly, we will use \({\sin ^2}t = \frac{{1 - \cos 2t}}{2}\)

Then the product of two functions of ‘t’ will be formed which will be further solved by

By parts formula: \(\int {uvdx = u\int {vdx - \int {\frac{{du}}{{dx}}\left( {\int {vdx} } \right)dx} } } \) will help to evaluate one given integral.

\( \Rightarrow \int {t{{\sin }^2}tdt} = \frac{1}{8}\left( {2{t^2} - 2t\sin 2t - \cos 2t} \right) + c\)

Let \(I = \int {t{{\sin }^2}tdt} \)** **

Now, as \({\sin ^2}t = \frac{{1 - \cos 2t}}{2}\)

\(I = \int {t\left( {\frac{{1 - \cos 2t}}{2}} \right)} dt\)

Now, split the integral as:

\(I = \frac{1}{2}\int {tdt - \frac{1}{2}\int {t\cos 2tdt \to (1)} } \)

The term can be integrated easily. For the II term we will use by parts formula

i.e., \(\int {uvdt = u\int {vdt - \int {\frac{{du}}{{dt}}\left( {\int {vdt} } \right)dt} } } \)

Let II term be denoted by \({I_2}\)

\(\begin{aligned}{l}{I_2} &= - \frac{1}{2}\int {t\cos 2tdt} \\ &= - \frac{1}{2}\left( {t\int {\cos 2tdt - \int {\left( {\frac{{dt}}{{dx}}\int {\cos 2tdt} } \right)} } } \right)dt\\ &= - \frac{{t\sin 2t}}{4} - \frac{{\cos 2t}}{8} + c \to (2)\end{aligned}\)

Using (2) in equation (1)

\(\begin{aligned}{l}I &= \frac{1}{2}\int {tdt - \frac{1}{2}\int {t\cos 2tdt} } \\I &= \frac{{{t^2}}}{4} - \frac{{t\sin 2t}}{4} - \frac{{\cos 2t}}{8} + c\\I &= \frac{{2{t^2} - 2t\sin 2t - \cos 2t}}{8} + c\end{aligned}\)

Hence, the value of integral:

\(\int {t{{\sin }^2}tdt} = \frac{{2{t^2} - 2t\sin 2t - \cos 2t}}{8} + c\)

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