pls help solve this q problem.

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Some Useful Formulas TABLE OF STANDARD INTEGRALS f (x) f(x) dx (1) ack, k * -1 pok+1 6+ 1 2 In | al 3 eat, a # 0 - par (4 sin x - COS x (5 COS D sin a (6 tan c - In cos x (7 cot x In sin x (8 sec T In secr + tanx (9 CSC - In csex + cot x (10) T a2 + 72: a20 a - tan ' (2 ) (11) Va? - x2, a > 0 Va2 - x2 - sin 1 (2) (12) Va2 +2, a > 0 Va2 + x2 + In a + val +x2 (13) Vx2 - a2, a > 0 Vx2 - a2 - In ac + Vac2 - a2 (14) 2 - 12:420 - In 2a for la| 0 sin -1 Va2 - 12 (16) a > 0 In Jac + vac2 + a2 Va2+ 12 (17) Vac2 - a >0 In a+vx2 -a2 (18 Inx (x > 0) x Inx - x (19) ear cos bx, a # 0, b > 0 par a cos br + b sin bax 2 + 62 ear sin bx, a # 0, b > 0 a sin bx - bcos bx (20) par 12 + 62Some reduction formulas: (1) f sin" xdx = -- cos r sin"-1 7 - [ sin"-2 x dx; n n (2) [ cost x dx = sin x cos"-1x n - + [ cos"-2 x dac; n n (3) [ tan" adr = tan"-Ix n - 1 + f tan"-2 x dx, (n > 1); (4) [ sec" edx = sec"-2 x tana n - 2 n - 1 J sec"-2 x d.x, (n > 1); (5) Sex" de = Learn-1 dx; a a2 (6) 1 ( x2 + a2 ) n -dx = 2n - 3 (2n - 2) (22 + a2 ) n-1 (2n - 2) (x2 + 92)n- 1de, (n > 1). Some trigonometric formulas: (1) sin(A + B) = sin A cos B + cos A sin B; (2) cos(A + B) = cos A cos B # sin A sin B; (3) sin 2A = 2 sin A cos B; (4) cos 2A = cos? A - sin A = 1 - 2 sin A = 2 cos2 A - 1; (5) sin A cos B = } [sin(A + B) + sin(A - B)]; (6) cos A cos B = ; [cos(A + B) + cos(A - B)]; (7) sin A sin B = } [cos(A - B) - cos( A + B)]