Integrate 3 2 t 3 − 5 2 t 2 + 2 t + cos ⁡ ( 3 t ) with respect to t.

Question

Integrate       3          2              t                  3                          ﻿    −      5          2              t                  2                      +  2  t  +  cos  ⁡  (  3  t  ) with respect to t.

rotgelb7kjxw · Accepted Answer

Split the single integral into multiple integrals.

$\int \frac{3}{2 t^{3}} d t + \int - \frac{5}{2 t^{2}} d t + \int 2 t d t + \int \cos (3 t) d t$

Since $\frac{3}{2}$ is constant with respect to $t$ , move $\frac{3}{2}$ out of the integral.

$\frac{3}{2} \int \frac{1}{t^{3}} d t + \int - \frac{5}{2 t^{2}} d t + \int 2 t d t + \int \cos (3 t) d t$

Apply basic rules of exponents.

$\frac{3}{2} \int t^{- 3} d t + \int - \frac{5}{2 t^{2}} d t + \int 2 t d t + \int \cos (3 t) d t$

By the Power Rule, the integral of $t^{- 3}$ with respect to $t$ is $- \frac{1}{2} t^{- 2}$ .

$\frac{3}{2} (- \frac{1}{2} t^{- 2} + C) + \int - \frac{5}{2 t^{2}} d t + \int 2 t d t + \int \cos (3 t) d t$

Since $- 1$ is constant with respect to $t$ , move $- 1$ out of the integral.

$\frac{3}{2} (- \frac{1}{2} t^{- 2} + C) - \int \frac{5}{2 t^{2}} d t + \int 2 t d t + \int \cos (3 t) d t$

Since $\frac{5}{2}$ is constant with respect to $t$ , move $\frac{5}{2}$ out of the integral.

$\frac{3}{2} (- \frac{1}{2} t^{- 2} + C) - (\frac{5}{2} \int \frac{1}{t^{2}} d t) + \int 2 t d t + \int \cos (3 t) d t$

Apply basic rules of exponents.

$\frac{3}{2} (- \frac{1}{2} t^{- 2} + C) - \frac{5}{2} \int t^{- 2} d t + \int 2 t d t + \int \cos (3 t) d t$

By the Power Rule, the integral of $t^{- 2}$ with respect to $t$ is $- t^{- 1}$ .

$\frac{3}{2} (- \frac{1}{2} t^{- 2} + C) - \frac{5}{2} (- t^{- 1} + C) + \int 2 t d t + \int \cos (3 t) d t$

Since $2$ is constant with respect to $t$ , move $2$ out of the integral.

$\frac{3}{2} (- \frac{1}{2} t^{- 2} + C) - \frac{5}{2} (- t^{- 1} + C) + 2 \int t d t + \int \cos (3 t) d t$

By the Power Rule, the integral of $t$ with respect to $t$ is $\frac{1}{2} t^{2}$ .

$\frac{3}{2} (- \frac{1}{2} t^{- 2} + C) - \frac{5}{2} (- t^{- 1} + C) + 2 (\frac{1}{2} t^{2} + C) + \int \cos (3 t) d t$

Let $u = 3 t$ . Then $d u = 3 d t$ , so $\frac{1}{3} d u = d t$ . Rewrite using $u$ and $d$ $u$ .

$\frac{3}{2} (- \frac{1}{2} t^{- 2} + C) - \frac{5}{2} (- t^{- 1} + C) + 2 (\frac{1}{2} t^{2} + C) + \int \cos (u) \frac{1}{3} d u$

Simplify.

$\frac{3}{2} (- \frac{1}{2} t^{- 2} + C) - \frac{5}{2} (- t^{- 1} + C) + 2 (\frac{t^{2}}{2} + C) + \int \frac{\cos (u)}{3} d u$

Since $\frac{1}{3}$ is constant with respect to $u$ , move $\frac{1}{3}$ out of the integral.

$\frac{3}{2} (- \frac{1}{2} t^{- 2} + C) - \frac{5}{2} (- t^{- 1} + C) + 2 (\frac{t^{2}}{2} + C) + \frac{1}{3} \int \cos (u) d u$

The integral of $\cos (u)$ with respect to $u$ is $\sin (u)$ .

$\frac{3}{2} (- \frac{1}{2} t^{- 2} + C) - \frac{5}{2} (- t^{- 1} + C) + 2 (\frac{t^{2}}{2} + C) + \frac{1}{3} (\sin (u) + C)$

Simplify.

$- \frac{3}{4 t^{2}} + \frac{5}{2 t} + t^{2} + \frac{1}{3} \sin (u) + C$

Replace all occurrences of $u$ with $3 t$ .

$- \frac{3}{4 t^{2}} + \frac{5}{2 t} + t^{2} + \frac{1}{3} \sin (3 t) + C$

Integrate 3 <mrow> 2 t 3 </mrow>

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