My fourth integration problem

Calculus Level 4

$\displaystyle \int_{0}^{{1} / {2} } \frac{dx}{\sqrt{9-36x^{2}-\arcsin^{2}(2x) +4x^{2} \arcsin^{2} (2x)}}$

If the integral above is equal to $\displaystyle \dfrac{a}{b} \arcsin \left( \dfrac{\arcsin(c)}{d}\right)$ , where $a,b,c$ and $d$ are positive integers, with $a, b$ are coprime, find $a+b+c+d$ .

The answer is 7.

1 solution

Rishabh Deep Singh
Jan 25, 2016

$\int _{ 0 }^{ 0.5 }{ \frac { dx }{ \sqrt { \left( { 3 }^{ 2 }-{ \left( \arcsin { 2x } \right) }^{ 2 } \right) \left( 1-{ \left( 2x \right) }^{ 2 } \right) } } } \\ now\quad put\quad \arcsin { 2x } =t\\ \quad \quad \quad \frac { dx }{ \sqrt { \left( 1-{ \left( 2x \right) }^{ 2 } \right) } } =\frac { dt }{ 2 } \\ \int _{ 0 }^{ \arcsin { 1 } }{ \frac { dt }{ 2\left( \sqrt { \left( { 3 }^{ 2 }-{ \left( t \right) }^{ 2 } \right) } \right) } } \\ =\frac { 1 }{ 2 } \arcsin { \left( \frac { \arcsin { \left( 1 \right) } }{ 3 } \right) }$

Moderator note:

What motivates the substitution of $t = \arcsin 2x$ ?

My fourth integration problem

The answer is 7.

1 solution

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