Strona 1 Edexcel Further Mathematics Further calculus Section 2: Inverse trigonometric functions Exercise level 2 solutions d 1 1. (i) (arctan x 2 )  2 x  dx 1  ( x 2 )2 2x  1  x4 d  1 1 1 (ii)  arcsin    2  dx  x x 1  ( x1 )2 ) 1  2 x 1  x12 1  x 2 1 x2 x2 1  x x2  1 d 1 1 (iii) (arccos(1  x ))   21 x 2   dx 1  (1  x )2 ) 1  2 x 1  (1  2 x  x ) 1  2 x 2 x x d 1 d (iv) (arcsin(arctan x ))   (arctan x ) dx 1  (arctan x )2 dx 1 1   1  (arctan x )2 1  x2 1  (1  x 2 ) 1  (arctan x )2 d 9 9 9 1  x  9   d d 2. (a)   d dx  1 4 d x   arctan     arctan  2 x  c 1  4x 2 c 1  x 4 2 4  21 1  2  c 2 c 9 9 9 dx  arctan  2 x 0  arctan 1  0 1/2 (i)  1/2 0 1  4x 2 2 2 9  8 1 of 3 16/10/18 © MEI integralmaths.org Strona 2 Edexcel FM Further calculus 2 Exercise solutions  9 9 dx  arctan  2 x  3 /2  (ii)  3 /2 1  4 x 2 2 9 9    2  arctan   arctan 3     2 2 3  3  4 d d 9 1 d 9  9  x  (b) c 1  4x dx   2 c dx   arcsin  1   2 4  x 2 2 1  2  c d  9    arcsin  2 x   2 c 1/2 1/2 9  9  9 (i)  0 1  4x 2 dx   arcsin  2 x   2 0  2 arcsin 1  0 9  4 3 /4 3 /4 9  9  (ii)  1 /2 2 1  4x 2 d x   2 arcsin  2 x    1 /2 2 9 3 1  9     arcsin  arcsin     2 2 2  2  3 4 3  8 3. (i)  arcsin 2 x dx   1  arcsin 2 x dx du 2 2 Let u  arcsin 2 x    dx 1  (2 x ) 2 1  4x 2 dv  1 v  x dx dv du Using integration by parts,  u dx  uv   v dx dx dx 2  1  arcsin 2 x dx  x arcsin 2 x   x  1  4 x 2 dx 2x  x arcsin 2 x   dx 1  4x 2  8 x(1  4 x 2  21  x arcsin 2 x  1 4 ) dx 1  x arcsin 2 x  41  2(1  4 x 2 )2  c  x arcsin 2 x  1 2 1  4x 2  c (ii)  arctan 3 x dx   1  arctan 3 x dx 2 of 3 16/10/18 © MEI integralmaths.org Strona 3 Edexcel FM Further calculus 2 Exercise solutions du 3 3 Let u  arctan 3 x    dx 1  (3 x )2 1  9x 2 dv  1 v  x dx dv du Using integration by parts,  u dx  uv   v dx dx dx 3x  arctan 3 x dx  x arctan 3 x   1  9 x 2 dx  x arctan 3 x  61 ln(1  9 x 2 )  c (iii)  1 (arcsin x  arctan x )d x  x x   x(arcsin x  arctan x )      dx  1x 2 1  x2  1  x(arcsin x  arctan x )  (1  x 2 )2  21 ln(1  x 2 )  c 3 of 3 16/10/18 © MEI integralmaths.org