Integrating Trigonometric Functions with Substitution

[karush]

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

Substituting

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

Which doesn't look like a good option ??

 

[Greg]

\(\displaystyle u=t^2+1,\quad\dfrac{du}{2}=t\,dt\)

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

Expand and integrate, then back-sub.

 

[karush]

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

 

[Prove It]

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 I meant the integral but I got it

Thanks the trig substitution would have been much harder

 

[Prove It]

Well I meant the integral but I got it

Thanks the trig substitution would have been much harder

Please stop editing your posts after you've received help, post in a new reply. We like to have the thread be coherent so that anyone who stops by having a similar problem can read through and get an idea what to do.

 

[karush]

In inclined more replies
https://dl.orangedox.com/GXEVNm73NxaGC9F7Cy