# [SOLVED]-w.8.7.16. Trig subst int

#### karush

##### Well-known member
$\tiny{8.7.16}$
$$\int{t}^{3}\sqrt{1+{t}^{2}} \ dr =\frac{\left({t}^{2}+1\right)^{5/2}}{5} -\frac{\left({t}^{2}+1\right)^{3/2}}{3}+C$$
$$t=\tan\left({u}\right) \ \ \ dt=\sec^2\left({u}\right) \ du$$

Substituting

$$\int\tan^3\left({u}\right)\sec^3\left({u}\right) \ du$$

Which doesn't look like a good option ??

#### Greg

##### Perseverance
Staff member
$$\displaystyle u=t^2+1,\quad\dfrac{du}{2}=t\,dt$$

$$\displaystyle \dfrac12\int(u-1)u^{1/2}\,du$$

Expand and integrate, then back-sub.

#### karush

##### Well-known member
$$\frac{1}{2}\int{u}^{3/2}-{u}^{1/2} \ du \implies \frac{1}{2} \left[\frac{2{u}^{5/2}}{5} -\frac{2{u}^{3/2}}{3}\right]+C$$
Thus
$$I=\frac{{\left({{t}^{2 }+1}^{}\right)}^{5/2}}{5} -\frac{{\left({{t}^{2 }+1}^{}\right)}^{3/2}}{3}+C$$

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#### Prove It

##### Well-known member
MHB Math Helper
I don't see how you got that
$$t={\left(u-1\right)}^{1/2}$$
\displaystyle \begin{align*} u &= t^2 + 1 \\ u - 1 &= t^2 \\ \left( u - 1 \right) ^{\frac{1}{2}} &= t \end{align*}

#### karush

##### Well-known member
Well I meant the integral but I got it

Thanks the trig substitution would have been much harder

#### Prove It

##### Well-known member
MHB Math Helper
Well I meant the integral but I got it

Thanks the trig substitution would have been much harder
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