Definite integral Problem 2

Hint: Substitute $x =(t+1)^2$ , then apply Integration by parts to get $a =\dfrac{16}{3}$ , and $b=3$

$\quad \quad \int _{ 1 }^{ 16 }{ \arctan { \sqrt { \sqrt { x } -1 } } dx } \\$

Put $x=\sec ^{ 4 }{ \theta } \\ \Rightarrow dx = 4\sec ^{ 4 }{ \theta } \tan { \theta }d\theta$

$\int _{ 0 }^{ \frac { \pi }{ 3 } }{ 4\theta \sec ^{ 4 }{ \theta } \tan { \theta } } d\theta$

Using product rule,

$\quad \quad \quad { 4\theta \int { \sec ^{ 2 }{ \theta } \tan { \theta } \sec ^{ 2 }{ \theta } d\theta } ) }-4\int _{ 0 }^{ \frac { \pi }{ 3 } }{ \int { \sec ^{ 2 }{ \theta } \tan { \theta } \sec ^{ 2 }{ \theta } d\theta } } d\theta \\ \Rightarrow \quad 4\theta \int { (\tan ^{ 2 }{ \theta } +1)\tan { \theta } \sec ^{ 2 }{ \theta } d\theta } )-4\int _{ 0 }^{ \frac { \pi }{ 3 } }{ \int { (\tan ^{ 2 }{ \theta } +1)\tan { \theta } \sec ^{ 2 }{ \theta } d\theta } } d\theta \\ \Rightarrow \quad 4\theta \left( \frac { \tan ^{ 4 }{ \theta } }{ 4 } +\frac { \tan ^{ 2 }{ \theta } }{ 2 } \right) -4\int _{ 0 }^{ \frac { \pi }{ 3 } }{ \left( \frac { \tan ^{ 4 }{ \theta } }{ 4 } +\frac { \tan ^{ 2 }{ \theta } }{ 2 } \right) } d\theta \\ \Rightarrow \quad \theta \tan ^{ 4 }{ \theta } +2\theta \tan ^{ 2 }{ \theta } -\int _{ 0 }^{ \frac { \pi }{ 3 } }{ (\tan ^{ 2 }{ \theta } ) } (\tan ^{ 2 }{ \theta } +2)d\theta \\ \Rightarrow \quad \theta \tan ^{ 4 }{ \theta } +2\theta \tan ^{ 2 }{ \theta } -\int _{ 0 }^{ \frac { \pi }{ 3 } }{ \tan ^{ 2 }{ \theta } } (\sec ^{ 2 }{ \theta } +1)d\theta \\ \Rightarrow \quad \theta \tan ^{ 4 }{ \theta } +2\theta \tan ^{ 2 }{ \theta } -\int _{ 0 }^{ \frac { \pi }{ 3 } }{ \tan ^{ 2 }{ \theta } \sec ^{ 2 }{ \theta } } d\theta -\int _{ 0 }^{ \frac { \pi }{ 3 } }{ \sec ^{ 2 }{ \theta } } d\theta +\int _{ 0 }^{ \frac { \pi }{ 3 } }{ d\theta } \\ \Rightarrow \quad \theta \tan ^{ 4 }{ \theta } +2\theta \tan ^{ 2 }{ \theta } -\frac { \tan ^{ 3 }{ \theta } }{ 3 } -\tan { \theta } +\theta$

Now, inserting limits,

$\frac { \pi }{ 3 } (9)+2\frac { \pi }{ 3 } (3)-\sqrt { 3 } -\sqrt { 3 } +\frac { \pi }{ 3 } -0\quad =\quad \frac { 16\pi }{ 3 } -2\sqrt { 3 } \\ a=\frac { 16 }{ 3 } ,\quad b=3$

And, you can do the math! :)

Let $p=\sqrt{\sqrt{x}-1}$ . Put $x=16$ to get $p=\sqrt{3}$ and $x=1$ to get $p=0$

Meanwhile, $x=\left(p^2+1\right)^2$ and $\mathrm dx=(4p^3+4p)\ \mathrm dp$ .

Now, we have new integration:

$I=\displaystyle \int_{0}^{\sqrt{3}} \left(4p^3+4p\right)\tan^{-1}p \ \mathrm dp$

We will apply integration by parts to solve $I$ .

Let,

$u=\tan^{-1}p$ , then $du=\frac{dp}{p^2+1}$

$v=\displaystyle \int 4p^3+4p \ \mathrm dp=p^4+2p^2$

$\begin{aligned}I&=\left(p^4+2p^2\right)\tan^{-1}p-\displaystyle \int{(p^2+1)}-\frac{1}{p^2+1} \ \mathrm dp\\ &= \left(p^2+1\right)^2\tan^{-1}p-\left(\frac{p^3}{3}+p\right) \end{aligned}$

Subtitution $p=\sqrt{3}$ and $p=0$ to find out the result of $I$ .

$I=16 \times\frac{\pi}{3}-2\sqrt{3}-0=\frac{16}{3} \times\pi-2\sqrt{3}$

Now, we just yield the required answer, $a\times b = \huge\boxed{16}$

For $\displaystyle{\int_{1}^{16}\ce{TAN} ^{-1}\sqrt{\sqrt{x}-1}\,dx}$

$\begin{aligned}\int_{1}^{16}\ce{TAN} ^{-1}\sqrt{\sqrt{x}-1}\,dx &= -\frac{1}{3}\left(\sqrt[3]{\sqrt{x} -1}\right)-\sqrt{\sqrt{x} -1}+x \ce{TAN}^{-1}\left( \sqrt{\sqrt{x}-1}\right)+\ce{\color{grey}{C}} \quad\quad\quad\quad\quad\quad\quad\quad{\text{Indefinite}}\\&= \frac{16}{3}\pi -2\sqrt{3} \end{aligned}$

where $a = \frac{16}{3}$ and $b= 3$

$\therefore \space a \cdot b = 16$ $\square$

$\large \text{ADIOSS!!!}$

X=sec^4@ and then by parts.

Hint: Substitute $x=(z^2+1)^2$ , then apply Integration by parts, we find $a=\frac{16}{3}$ and $b=3$ .

So, $ab=\boxed{16}$