Question 5

x log 2x 6. x2 log x 7. x sin- 1 x 8. x tan -1 x 9. x cos -1 x 10. (sin-1 x)2 11. 1/2 cos x x x − − 12. x sec 2 x 13. tan -1 x 14. x (log x)2 15. (x2 + 1) log x 16. ex (sinx + cosx) 17. 2(1 ) xx e x+ 18. 1 sin 1 cos x xe x  +  +  19. 2 1 1-xe x x      20. 3 ( 3) ( 1) xx e x − − 21. e2x sin x 22. 1/2 2sin - x x    +  Choose the correct answer in Exercises 23 and 24. 23. 32 xx e dx∫ equals (A) / 31 C3 / xe + (B) 21 C3 xe + (C) 31 C2 / xe + (D) 21 C2 xe + 24. sec (1 tan )xe x x dx +∫ equals (A) ex cos x + C (B) ex sec x + C (C) ex sin x + C (D) ex tan x + C 7.6.2 Integrals of some more types Here, we discuss some special types of standard integrals based on the technique of integration by parts : (i) 2 2x a dx−∫ (ii) 2 2x a dx+∫ (iii) 2 2a x dx−∫ (i) Let 2 2I x a dx= −∫ Taking constant function 1 as the second function and integrating by parts, we have I = 2 2 2 2 1 2 xx x a x dx x a − − −∫ = 2 2 2 2 xx x a dx x a − − −∫ = 2 2 2 2 2 2 2 x a ax x a dx x a − +− − −∫ = 2 2 2 2 2 2 2 dxx x a x a dx a x a − − − − −∫ ∫ = 2 2 2 2 2

x a / − − − / −∫ or 2I = 2 2 2 2 2

x a / − − / −∫ or I = ∫ 2 2x - a dx = 2 2 2 2- - log + - + C2 2

Similarly, integrating other two integrals by parts, taking constant function 1 as the second function, we get (ii) ∫ 2 2 2 2 2 21+ = + + log + + + C2 2

(iii) Alternatively, integrals (i), (ii) and (iii) can also be found by making trigonometric substitution x = a secθ in (i), x = a tanθ in (ii) and x = a sinθ in (iii) respectively. Example 23 Find 2 2 5x x dx+ +∫ Solution Note that 2 2 5x x dx+ +∫ = 2( 1) 4x dx+ +∫ Put x + 1 = y, so that dx = dy. Then 2 2 5x x dx+ +∫ = 2 2 2y dy+∫ = 2 21 44 log 4 C2 2y y y y+ + + + + [using 7.6.2 (ii)] = 2 21 ( 1) 2 5 2 log 1 2 5 C2 x x x x x x+ + + + + + + + + Example 24 Find 23 2 x x dx− −∫ Solution Note that 2 23 2 4 ( 1)x x dx x dx− − = − +∫ ∫ Put x + 1 = y so that dx = dy. Thus 23 2 x x dx− −∫ = 24 y dy−∫ = 2 11 44 sin C2 2 2 - yy y − + + [using 7.6.2 (iii)] = 2 11 1( 1) 3 2 2 sin C2 2 - xx x x + + − − + +  