• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q11E

Expert-verified
Found in: Page 326

### Essential Calculus: Early Transcendentals

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.

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

See the step by step solution

## Step-1:  Using trigonometric double angle formula: -

Let $$I = \int {t{{\sin }^2}tdt}$$

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

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

## Step-2:  Splitting of integral

Now, split the integral as:

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

## Step-3:  Using by parts integration formula

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}

## Step-4:  Using $${I_2}$$ in $$I$$

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:

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