arcsin-x-arccos-x-dx-

Question Number 52898 by MJS last updated on 14/Jan/19

\int \arcsin x \arccos x d x = ?

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

a = {s i n}^{-} x

s i n a = x

{s i n}^{- 1} x + {c o s}^{- 1} x = \frac{π}{2}

\int a \times (\frac{π}{2} - a) \times c o s a d a

\frac{π}{2} \int a c o s a d a - \int a^{2} c o s a d a

I = \frac{π}{2} I_{1} - I_{2}

I_{1} = \int a c o s a d a I_{2} = \int a^{2} c o s a d a

I_{1} = a \int c o s a d a - \int [\frac{d a}{d a} \int c o s a d a] d a

= a s i n a - \int s i n a d a

= a s i n a + c o s a + c_{1} \leftarrow v a l u e o f I_{1}

I_{2} = \int a^{2} c o s a d a

= a^{2} s i n a - \int 2 a s i n a d a

= a^{2} s i n a - I_{3}

I_{3} = \int 2 a s i n a d a

= 2 a (- c o s a) - \int 2 \times - c o s a d a

= - 2 a c o s a + 2 s i n a

s o I_{2} = a^{2} s i n a + 2 a c o s a - 2 s i n a

S o I = \frac{π}{2} I_{1} - I_{2}

= \frac{π}{2} (a s i n a + c o s a) - (a^{2} s i n a + 2 a c o s a - 2 s i n a) + c

= \frac{π}{2} ({x s i n}^{- 1} x + \sqrt{1 - x^{2}}) - [{({s i n}^{- 1} x)}^{2} x + 2 {s i n}^{- 1} x \times \sqrt{1 - x^{2}} - 2 x) + c

s i r p l s c h e c k \dots