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

Question

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

hodowlanyb1rq2 · Accepted Answer

Split the single integral into multiple integrals.

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

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

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

By the Power Rule, the integral of $t^{3}$ with respect to $t$ is $\frac{1}{4} t^{4}$ .

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

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

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

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

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

The integral of $\frac{1}{t}$ with respect to $t$ is $\ln (| t |)$ .

$\frac{3}{2} (\frac{1}{4} t^{4} + C) - \frac{5}{2} (\ln (| t |) + C) + \int \sin (2 t) d t + \int 2 \cos (2 t) d t$

Let $u_{1} = 2 t$ . Then $d u_{1} = 2 d t$ , so $\frac{1}{2} d u_{1} = d t$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

$\frac{3}{2} (\frac{1}{4} t^{4} + C) - \frac{5}{2} (\ln (| t |) + C) + \int \sin (u_{1}) \frac{1}{2} d u_{1} + \int 2 \cos (2 t) d t$

Combine $\sin (u_{1})$ and $\frac{1}{2}$ .

$\frac{3}{2} (\frac{1}{4} t^{4} + C) - \frac{5}{2} (\ln (| t |) + C) + \int \frac{\sin (u_{1})}{2} d u_{1} + \int 2 \cos (2 t) d t$

Since $\frac{1}{2}$ is constant with respect to $u_{1}$ , move $\frac{1}{2}$ out of the integral.

$\frac{3}{2} (\frac{1}{4} t^{4} + C) - \frac{5}{2} (\ln (| t |) + C) + \frac{1}{2} \int \sin (u_{1}) d u_{1} + \int 2 \cos (2 t) d t$

The integral of $\sin (u_{1})$ with respect to $u_{1}$ is $- \cos (u_{1})$ .

$\frac{3}{2} (\frac{1}{4} t^{4} + C) - \frac{5}{2} (\ln (| t |) + C) + \frac{1}{2} (- \cos (u_{1}) + C) + \int 2 \cos (2 t) d t$

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

$\frac{3}{2} (\frac{1}{4} t^{4} + C) - \frac{5}{2} (\ln (| t |) + C) + \frac{1}{2} (- \cos (u_{1}) + C) + 2 \int \cos (2 t) d t$

Let $u_{2} = 2 t$ . Then $d u_{2} = 2 d t$ , so $\frac{1}{2} d u_{2} = d t$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

$\frac{3}{2} (\frac{1}{4} t^{4} + C) - \frac{5}{2} (\ln (| t |) + C) + \frac{1}{2} (- \cos (u_{1}) + C) + 2 \int \cos (u_{2}) \frac{1}{2} d u_{2}$

Combine $\cos (u_{2})$ and $\frac{1}{2}$ .

$\frac{3}{2} (\frac{1}{4} t^{4} + C) - \frac{5}{2} (\ln (| t |) + C) + \frac{1}{2} (- \cos (u_{1}) + C) + 2 \int \frac{\cos (u_{2})}{2} d u_{2}$

Since $\frac{1}{2}$ is constant with respect to $u_{2}$ , move $\frac{1}{2}$ out of the integral.

$\frac{3}{2} (\frac{1}{4} t^{4} + C) - \frac{5}{2} (\ln (| t |) + C) + \frac{1}{2} (- \cos (u_{1}) + C) + 2 (\frac{1}{2} \int \cos (u_{2}) d u_{2})$

Simplify.

$\frac{3}{2} (\frac{1}{4} t^{4} + C) - \frac{5}{2} (\ln (| t |) + C) + \frac{1}{2} (- \cos (u_{1}) + C) + \int \cos (u_{2}) d u_{2}$

The integral of $\cos (u_{2})$ with respect to $u_{2}$ is $\sin (u_{2})$ .

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

Simplify.

$\frac{3 t^{4}}{8} - \frac{5 \ln (| t |)}{2} - \frac{\cos (u_{1})}{2} + \sin (u_{2}) + C$

Substitute back in for each integration substitution variable.

$\frac{3 t^{4}}{8} - \frac{5 \ln (| t |)}{2} - \frac{\cos (2 t)}{2} + \sin (2 t) + C$

Reorder terms.

$\frac{3}{8} t^{4} - \frac{5}{2} \ln (| t |) - \frac{1}{2} \cos (2 t) + \sin (2 t) + C$

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