Find the indefinite integral $\int_{} \frac{t^3 dt}{\sqrt{1+t^2}}$

Question

Find the indefinite integral $\int_{} \frac{t^3 dt}{\sqrt{1+t^2}}$

1) Choose a u

I decide to let $$u = 1 + t^2, du = 2t$$

2) Substitute u value into integral
$$\int_{} \frac{t^3}{\sqrt u} du$$

3) Integrate
$$\int_{} \frac{t^3}{\sqrt u} du = \frac{t^4}{4} \frac{u^{1/2}}{1/2} \frac{2t^2}{2} + C$$

4) Put substitutes into anti-derivative $$= \frac{t^4}{4} 2(1+t^2)^{1/2} + C = \frac{1}{2}t^4(1+t^2)^{1/2} + C$$

$~$ $~$

However, the answer for this problem is $$\frac{1}{3}(1+t^2)^{3/2} - (1 + t^2)^{1/2} + C$$

Where did I go wrong? My answer doesn't even look close, except for the fact that I chose the right $u$ It looks like I didn't fully evaluate for the anti-derivative or I incorrectly set the problem up before evaluating for the anti-derivative (I think this it is the latter)

Answer 1

Use $u = 1+t^2$ and you find $\mathrm{d}u = 2t$. So $$t^2 = u-1 \\ t^3 = \frac{1}{2}(u-1)\:\mathrm{d}u$$ Your integral will be $$\frac{1}{2}\int\frac{u-1}{\sqrt{u}}du = \frac{1}{2}\left(\int\sqrt{u}\mathrm\:{d}u - \int \frac{1}{\sqrt{u}} \: \mathrm{d}u\right)$$

Now integrate and substitute for $u$ to find the desired answer.

You will find the answer to be $$\frac{1}{3}u^{\frac{3}{2}}-u^{\frac{1}{2}}+ C$$

Answer 2

Hint: Substitute $$u=\sqrt{1+t^2}$$ then we get $$u^2-1=t^2$$ and $$udu=tdt$$ so $$dt=\frac{u}{t}du$$ and our integral will be $$\int\frac{t^2u}{u}du=\int u^2-1du$$

Answer 3

$$I=\int\frac{x^3dx}{\sqrt{1-x^2}}=\int\frac{x^2}{\sqrt{1-x^2}}xdx$$ Substitution: $$u=1-x^2\Rightarrow -\frac{du}{2}=xdx$$ So $$I=-\frac12\int\frac{1-u}{\sqrt{u}}du$$ $$I=\frac12\int\frac{u-1}{\sqrt{u}}du$$ $$I=\frac12\int\frac{u}{\sqrt{u}}du-\frac12\int\frac{du}{\sqrt{u}}$$ $$I=\frac12\int\sqrt{u}du-\frac12\int\frac{du}{\sqrt{u}}$$ $$I=\frac13u^{3/2}-u^{1/2}$$ $$I=\frac13(1-x^2)^{3/2}-(1-x^2)^{1/2}+C$$ $$I=\frac12(1-x^2)^{1/2}\left(x^2-\frac13\right)+C$$