Integrate 1 2 t − 5 2 sin ⁡ ( 2 t ) + 1 2 sin ⁡ ( 3 t ) − 2 with respect to t.

Question

Integrate       1          2      t        −      5    2    sin  ⁡  (  2  t  )  +      1    2    sin  ⁡  (  3  t  )  −  2 with respect to t.

Semaj Stark · Accepted Answer

Split the single integral into multiple integrals.

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

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

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

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

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

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

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

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 $d$ $u_{1}$ .

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

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

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

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

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

Simplify.

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

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

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

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

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

Let $u_{2} = 3 t$ . Then $d u_{2} = 3 d t$ , so $\frac{1}{3} d u_{2} = d t$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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

Combine $\sin (u_{2})$ and $\frac{1}{3}$ .

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

Since $\frac{1}{3}$ is constant with respect to $u_{2}$ , move $\frac{1}{3}$ out of the integral.

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

Simplify.

$\frac{1}{2} (\ln (| t |) + C) - \frac{5}{4} (- \cos (u_{1}) + C) + \frac{1}{6} \int \sin (u_{2}) d u_{2} + \int - 2 d t$

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

$\frac{1}{2} (\ln (| t |) + C) - \frac{5}{4} (- \cos (u_{1}) + C) + \frac{1}{6} (- \cos (u_{2}) + C) + \int - 2 d t$

Apply the constant rule.

$\frac{1}{2} (\ln (| t |) + C) - \frac{5}{4} (- \cos (u_{1}) + C) + \frac{1}{6} (- \cos (u_{2}) + C) - 2 t + C$

Simplify.

$\frac{\ln (| t |)}{2} + \frac{5 \cos (u_{1})}{4} - \frac{\cos (u_{2})}{6} - 2 t + C$

Substitute back in for each integration substitution variable.

$\frac{\ln (| t |)}{2} + \frac{5 \cos (2 t)}{4} - \frac{\cos (3 t)}{6} - 2 t + C$

Reorder terms.

$\frac{1}{2} \ln (| t |) + \frac{5}{4} \cos (2 t) - \frac{1}{6} \cos (3 t) - 2 t + C$

Integrate 1 <mrow> 2 t </mrow> − 5 2 sin &#x206