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Indefinite Integration(IIT-JEE)

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Published in: IIT JEE Mains
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Dear Students, Solve every problem without taking the help of solution. At the end you can see the solutions to improve yourself.

Rakesh K / Delhi

7 years of teaching experience

Qualification: B.Tech/B.E. (Galgotias College of Engineering & Technology (GCET), Greater Noida - 2013)

Teaches: Business Mathematics, Computer Science, IT & Computer Subjects, Mathematics, Physics, KVPY Exam, PSU Exam, AIEEE, BITSAT, GATE Exam, IIT JAM, IIT JEE Advanced, IIT JEE Mains, Computer, Electronics And Communication, C / C++

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  1. 1 = f (x) g (x). then g (x) can be 1 x C Differentiating on both sides w. r. to x, we get g(x) = 2. C 2. (x2 +cos2 x) . cos ecx.dx 2 (1 +x2 —sin2 x) . cos ec2 x.dx 2 cos ec2 x.dx — cot x — tan—I x + c tan- x (1 + x + d (cot-Ix) is equal to 3. Je (A) - etan x (C) — X. etan x 3. 4. (A) Sln 4. + cos2 x . cosec x dx is equal to 2 (A) cot x + tan-Ix + c (C) — cot x — tan-Ix + c (B) cot x — tan-Ix + c (D) tan-Ix — cot x + c tan x (D) X. etan x C tan x Let I = fe 1 dx 2 tan tan tan- x Xdx Xdx tan x dx 2 tan x dx 2 tan- x tan- x xe x x.e x (ax2 -b)dx is c2x2 —(ax2 + ax + b/ x C ax + b/ x (C) cos c ax (B) sin ax (D) cos +b/x2 c +b/x2 c
  2. The value of (??2 ? ?? Let = dx = b)dx ?? +?/? 1 ?2?) dx 2n ? ??? 1 is 1 1 = sin-1 (?) 1 ? 2n ? (1 - ?2?) dx (0) 1-?? ? 1— ?2? ??? 1 2? 1/2 (1 -??) 1-?2? = dx = (1 ??) 1—?2? (1 ? cot ? cos ??? The value of dx is equal to (? + cot (xsinx + cosx) (?) (? + cot ?) (xtanx + 1) (?) ???? of these ? cot ? cos ??? dx = (? + cot Let ? sin ? + cos ? = t ? cos ? dx = dt ? cot ? cos ??? ??? ?(? tanx +1) xcosx dx (xsinx + cos dt The value of (?) ? (3?28 84 — (3?21 80 (3?21 + 7 ? (xsinx+ cosx) (3?21 4-7 ? + 7?12 +21?4 + 7?9 +21? 1/3 +21? 20 + 21?3 ? 20 + ? 8 + ? put 3?21 + 7?9 + 21?3 = t 63[?20 + + x2]dx = dt 4/3 l2 - t3dt 63 4 63 —(??21 84 — (3?20 + 7?8 + 21? 21 (?) None of these 4/3 + 21?3
  3. 84 The value of Isinxlog cot— dx = (C) cos x log cot — cos x log tan— x is log sin — + log — log sin— log sec cos ec cot I f cosx dx = sin x log cot— + log tan— + c sin x log cot— +cosxlog tan cosx)dx sinxlog cot— cosxlog cot— cosxlog cot— 2 cosxlog cot— 2 cosxlog cot— 2 cosxlog cot— The value of cos sin cos x log cot — 2 dx 72 J cos x sin fcot—dx 2 210g sin— 2 72 J cos—sin f tan—dx 210g sec— log sin— +log sea +c sinx sin log tan x x sin x + + log tan 2 12 x X dx is 12 log tan —+— + log tan 2 12 2 logcot —+— 2 12 (D) None of these (where c is the integration constant) sinx sin 2 sin x cosit/6 sin x sin x dx sin x +
  4. sin +Sin x 1 6 6 dx sin x sin x + 6 6 1 cosec x + cos 6 1 log tan + log tan 2 12 6 2 12 10. 10. 11. 11. 12. where c is the integration constant. 1 The value of dx is x 2 2+3x 2 1 x where c is integration costant 1 dx 1 dx Let x - -JfüTdt = 1 x 1/3 3 3 x -1/3 2 2 2+3x cos ec2x - 2005 dx is equal to 2005 cos x cot x 2005 (cos x) (tan x) (C) 2005 (cos x) cos ec2x - 2005 dx - 2005 cos x -2005 (cos x) • cot x) 4 x x 2 2+3x (D) None of these tan x (B) 2005 (cos x) (D) none of these - f(cosx) -2005 cos ec f (-2005) (cosx) -2006 is - 20051 2005 x sin x)(—cot x) dx — 20051 2005 x cot x 2005 (cos x) (ax2 b)dx x
  5. ax + b/ x (A) Sin I c ax + b/ x ax (B) Sin I ax (D) COS I 12. dx= Sin-I 13. 13. 14. 14. 15. 15. 16. (C) COS I (ax2 x c2x2 c b)dx 1 b 5 +b/x2 c +b/x2 c b 1 + (sin d (sin 1 + (sin 1 (sin)2/3 1 a 2 c x 2 ax + b/ x 1 c is equal to 1 1 (B) —tan x c -l (sin)2/3 1 1 tan B dx (A) sec I + c -1 (C) —sec c (sin)l/3 é(sin is equal to cos5x + cos 4x dx is equal to 1-2cos3x (A) — sin 2x — sin x + c sin 2x (D) none of these (B) 2 sec I + c (D) none of these sin2x (B) sinx+c 2 (D) none of these (C) + sin x + c 2 cos5x + cos 4x dx 1-2cos3x 9x (cos 5x + cos 4x) sin3x dx sin3x — sin6x 2cos cos sin3x 2 2 dx 9x 3x 2cos sin 2 2 sin 2x sinx+c. 2 COS x + xsinx dx is equal to x(x + cos x) (cos 2x + cos x)dx
  6. ? (?) log ? + cosx xsinx (?) log ? + cos ? 6 ? + cosx (?) log ? (?) ???? of these cosx+xsinx ? + cosx—X+ xsinx dx ?(? + cos ?) 1—sinx dx ?(? + cos ?) ? dx = log ? + cosx ? + cos ?