integration_JEE MAINS_objective

本站添加: 2026-05-19

[00:02:16][clears throat] So [clears throat] till now we discussed about integration by substitution. So now integration by parts integration by parts.

[00:03:05]Here we know that dx of u v is equal to u v dash + v u dash that is u dy dx of v plus dy dx of right.

[00:03:33]So dy dx of fx into gx of x intox. So how we can write this?

[00:03:51]FX into a by dx of gx plus gx into a dx of fx.

[00:04:05]So this is derivative of product of fx and gx. So what about it anti-derivative integral fx into dy dx of gx plus gx into dy dx of f_sub_x dx is equal to integral d by dx of f_sub_x into gx dx.

[00:04:49]Okay. So here integral dative antiative derative gets cancel. So this is all fx= fx into gx. Here it is it is f_sub_1 it is f_sub_2 both combined by fundamental operation addition. So we can split the integral for both. So integral fx into d by dx of gx integral uh dx plus integral gx into d by dx of f_sub_x dx. That is all equal to f(x) into letter f_sub_x = u d by dx of gx = d by dx of gx = so that implies integral dy dx of gx dx = integral dx integral dx. So that implies dy dx of gx dx integral integral and derivative antiative gets cancel integral v dx. So what is the assumption? gx is integral v dx and fx is right. So now these going to be substituted here integral what about fx u what about dy dx of dx dx integral u v dx is equal to what is fx u what is dx integral >> this plus one side minus integral f(x) what is f(x) u. So d by dx of u into what is dx?

[00:07:18]This is integral u dx derivation. So what is the formula for u dx?

[00:07:32]U integral V dx >> V integral integral PX of U integral V dx whole dx that is so here the thing is who is v who is u first function second function.

[00:08:06]So who is we? Who is you? So we can identify by taking this is the order.

[00:08:17]So I for inverse triome, L for square, A for all, P for trigonometry, E for exponential.

[00:08:42]So this is the order for this is the order for.

[00:08:52]So for example integral sin roo<unk>x dx integral sin <unk>x dx. So how can we solve this?

[00:09:08]Let xx = t 1 by 2<unk>x dx = d. So dx = 2t dt because x is so integral dx 2t dt sin <unk>x dt.

[00:09:35]So here it is in order all the trigonometric. This is first function.

[00:09:42]This is second function. Right?

[00:09:47]This is u. This is v. So uh u integral 2 into u integral sin t dt minus integral d by dx of t into sin t dt.

[00:10:15]Okay. So what about sin t minus 2 into t into minus cos t minus what about dy dt sorry dtus t this becomes plus cos d so that is equal to 2 into minus E cos t plus >> sin t plus c. So this can be written as 2 into -<unk> x cos x + sin<unk> x >> [cough and clears throat] >> That's not So the next application e x into fx + f x = e x fx + 6.

[00:18:51]So integral e power x fx dx plus integral e power x of x dx.

[00:19:03]So let it be 1 to [clears throat] u v. So, e power x integral f x as per exponential consider as v. So this is u this is v.

[00:19:36]So f(x) integral e x dxus integral d by dx of fx into integral f_sub_x dx whole dx plus this what is e xd of x dx.

[00:19:58]So now this is integral.

[00:20:16]So e x integration e x fxus this is fx and dx integral x dx frx plus e x integral frx.

[00:20:36]So these two gets cancel.

[00:20:40]So e power x for example integral e power x into tan x +² x dx.

[00:20:55]>> So this is fx this is fx. So this is equal to e x tan x Integral E power X into 1 + sin 2X by 1 + cos 2x Yes, Richard.

[00:24:35]both ways. Sorry.

[00:25:53]So integral e^ 2x into 1 + 2 sin x cos x by dx.

[00:26:07]Okay. Here 2x = p dx = 1x 2 dt x = 2 [clears throat and cough] h.

[00:26:23]So 1x2 integral e power t uh what about here 1x2 see square plus 2 sin 2 cos x 2 cos x sin x means tan right so if you observe if it is fx X it is >> center of X. So 1 by 2 e power tan t plus = 1x 2 e 2x tan x.

[00:27:23]No question.

[00:27:27]Integral log of log x + 1 by log x² dx.

[00:27:49]Next question.

[00:29:46]First log x is equal to log x base = d. x = e power d 1 by x dx = dt dx = e power dx = e dt.

[00:30:20]So integral log d plus 1 t² e power d right. So what is log dative?

[00:30:40]Yeah, there is no 1x. So add and subtract 1x integral log t + 1 by t - 1 by t + 1x t² [clears throat] integral log 3 c - 1 by t plus 1x T + 1 by T² + 1X² + 1 by So this is FX this is so E power t into c minus what is t.

[00:31:50]So ei means x into log of log x c - 1 by log x >> [clears throat and cough] >> Integral EX sin dx + c x sin x + That will be I I = E power X integral sin PX + C dx - d by dx of e x integral sin px + c dx dx.

[00:36:18]So e x this is - cos bx + c by bin dx + c dx = d. So dx is equal to minus integral e x a e power a x into - cos bx + c by b dx right so i = - 1 by b e a x cos bx + C - A by B - plus A by B integral E power A X into cos BX + C dxus integral d by dx of e power ax into integral cos dx + c dx whole dx.

[00:37:43]So I = - 1 by e power a x cos bx + c plus a by b e power a x cos b x + c by b minus m* b a by b into again x of x a= A s into this is all under integration.

[00:38:20]Integral Ax into sin B X + C by B X - B by A E power A X cos B X + C plus this is A by B² E power A X cos B sin BX x + c - a b a by b a² by b². What about e x sin bx + c dx? e x sin bx + c dx i. So this is replaced by i. of this term transpose to LHS plus a² by² I is equal to what is LCM here b² so a e power a x sin b x + c - b e a x cos b x + Right.

[00:39:49]I into A² + B² B² = E X A sin B X + C - B cos B X + C by square B square so I = a X sin B X + C dx is equal to what is the result here?

[00:40:25]A square + B² mean E by A² + B² into A B X + C - B cos So after this second e X cos Bx + 6 by A² + B² A cos B + C + B sin B X Thanks.

[00:42:26][cough] [clears throat] >> [snorts] >> Let's just Uh, that's it. That's it.

[00:45:08]Uh, Uh hey number PN same process to be complete squaring. So 7 x - 10 - x². So - common x² - 2 into x into 7x2 + 7 by 2² - 7x2² + 10us of x - 7x2² + 10 - 49 by 4 x² square - 2 into x into 7 by2 gets cancel 7 x + 10 b² - b² right so this is 40 - 49 - 9 - of minus so this is 3x 2 - x - 7x 2 so<unk> over uh 3x X2² - X - 7X 2² dx. So for a square x2 x2 x - 7 by 2 by 2 into roo<unk> a² - x² roo<unk> a square - x² x - 10 - X² + A² by 2 A² 9 by 4 by 2 sin inverse uh X - 7 by 2 by 3X 2x - 7 by 4<unk> 7 x - 10 - 9 by 8 sin inverse 2x - 7 by 3.

[01:02:57]Okay.

[01:03:01]Simplic.

[01:03:13]important for Is there integral power x² square + b x + c dx = integral / a into x + b / 2 a² + 4 a c - b² by 4 axal dx + u into roo<unk> / x² + b x + c dx px + q into<unk> x² + model. So here px + q can be written as a intox of a x² + b x + coo b.

[01:07:20]Okay. Example integral x + 1<unk> / x + 1<unk> / 1 - x - x² So, x + 1 = a into dx of 1 - x - x² + b.

[01:08:36]So, x + 1 = a into - 1 - 2x 1 x 1 x 2x + b. So compare the coefficients - 2 a = 1 here that implies a = - a + b = 1 = 1 + a that is equal to 1 by 2. So now integral x + 1<unk> 1 - x - x² dx can be written as a value - 1 by 2 integral. What is the derative of this one?

[01:09:30]Uh -1 - 2x<unk> 1 - x - x² dx + 1x 2 integral<unk> 1 - x - x² dx 1 - x - x² - 1 - 2x dx = dt So - 1 by 2 integral<unk> t dt roo<unk> t power of dt + 1 by 2 integral<unk> /us common x² - 2 into + 2 into x into 1 by 2 + 1x 4 - 1 by 4 - 1 dx.

[01:10:36]So - 1 by 2 into t power 3x2 by 3x2 x + 1 by 1 by 2 integral<unk> / <unk>5 by 2² - x + 1 by 2² x right here 2x 2x cancel - 1x 3 into what is x 1 - x - x² 3x2 + 1 by 2 into a square - x² a square - x² what is formula xx 2 x + 1 by 2 sorry x + 1 by 2 by 2 into roo<unk> 1 - x - x² + a² okay 5x 4 by 2 sin inverse x by A x + 2 by<unk> 5 by sorry x + 1 by 2 by<unk> 5 by Hey, Next topic partial fractions.

[01:16:39]Two state integration.

[01:16:46]So first of all we have to know about partial fractions.

[01:16:54]Let f of x g of x be two polomials such f of x degree less than or equal to g of x degree.

[01:17:19]f of x by g of x f of x by >> g of x one g of x = x - alpha 1 x - alpha 2 and so on x - alpha n which are linear factors All are linear factors. Degree one.

[01:17:54]Case one.

[01:17:56]Linear factors.

[01:18:08]All are non repeated.

[01:18:15]linear factors.

[01:18:19]H long repeated linear factors f_sub_x by gx = a1 by x - ala 1 + a2 by x - ala 2 and so on a n by x - ala to G of X has linear factors.

[01:19:04]Some are repeated, rest are non repeated.

[01:19:16]Some are repeated, the remaining are non repeated. So suppose gx = for example x - alpha 1² into x - ala 2 cube into x - ala 3 x - alpha Right? So these two are repeated. The remaining two are nonre repeated otherwise.

[01:20:02]So now f_sub_x by zx = a1 by x - ala 1 + a2 by x - ala 1 + a3 by x - ala 2 + a4 by x - ala 3 + a5 by x - ala 4 plus a 6 by x - ala 4² square + a 7 by - ala 4.

[01:20:44]Next case three.

[01:20:47]G of X has some linear repeated and long repeated.

[01:21:10]and some irreducible quadratic factors.

[01:21:31]Reducable quadratic factors g of x = a x² + bx + c into x - ala 1 into x - 2 fx by gx f_sub_x by gx = A1 x + a2 by a x² + b x + a3 x + a4 example And foremost condition is f of x degree less than or equal to g of x less than not less than or equal right.

[01:22:41]If it is equal if it is more then also same process that if it is more or equal we are going to take division. For example, x^ 4 by x + 1 x + 1 x^ 4 x into x x^ 4 + x.

[01:23:17]So uh this is - x - x by x + 1.

[01:23:23]So integration for this one and it is to be factorization prime facial fraction. So see x by xq + 1 a by x + 1 + b x + c by x² - x + 1 x + x + 1 x² - x + x + 1 factors a + b a + b into a square - a + b² so this is quadratic As for third quadratic vineg integration Class evening 6 to 8 fractions integration models complete and then exercise concepts.

[01:25:03]So session concludes Again split into two linear cannot be factorized.