Integrate 2 t 4 − 3 2 t 3 + t − 3 sin ⁡ ( t ) 2 with respect to t.

Question

Integrate       2          t              4              −      3          2              t                  3                      +  t  −            3      sin      ⁡      (      t      )        2   with respect to t.

priffEmipsy4i37m · Accepted Answer

Split the single integral into multiple integrals.

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

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

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

Apply basic rules of exponents.

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

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

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

Simplify.

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

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

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

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

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

Apply basic rules of exponents.

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

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

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

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

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

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

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

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

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

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

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

Simplify.

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

Integrate 2 t 4 </mrow> − 3 <mrow>

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