INTEGRATION_JEE MAINS_CLASS X_FOUNDATIONS

Added: 2026-05-11

[00:04:17]So integral 1x sin x - a into sin x - b 1 by sin a - b into sin a minus b or else it is b minus a we take sin x - a into sin x - Okay.

[00:04:59]Nothing but partial fractions 1 by sin b - a.

[00:05:07]So now this can be written as x - a - x - b by sin x - a x - b dx b - a add and subtract x minus of minus + b right now this is in the form of sin a minus b 1x sin b - a into sin a minus b sin a minus b sin a cos b minus cos a sin b by sin x - a sin x - b dx So if you'll split into fractions 1x sin b - a integral c this one by this one >> sin x - a sin x - a gets cancel cos x - b by >> sin x - dx X - sin X - B sin X - B gets cancel integral cos X - A by sin X - A dx right 1 by sin B - A into F of X by f(x) dxing.

[00:07:11]mod fx right sin x - b - log sin x - Okay. So 1 sin b - a into log - log m sin x - b by sin x - a modus.

[00:08:01]Got it.

[00:08:03]Okay.

[00:08:51]In the same way you have to solve it.

[00:09:04]Integral 1 by cos x - a into cos x - b dx.

[00:10:38]Here also you have to consider same 1x sin bip for this also numerator become sin x - y sin x cos y - cos x sin y then simplify 1x sin b - a cos x log cos x - a by cos That's the required answer.

[00:13:49]Okay.

[00:13:55]Integral tan x tan 2x tan 3x dx.

[00:14:04]Integral tan x tan 2x tan 3x dx.

[00:14:14]that you have to serve.

[00:16:02]You know tan a + b is equal = tan a + tan b + tan a + b tan a tan b this is a trigonometric application tan a + b horizontal formula tan a + tan b + tan a + b into tan a tan b so let a = 2x X B = X then what we get here from starts from here onwards tan A + B 3X tan 2x tan x is equal to tan a + b tan 3x these two in this side tan 2x - >> tan x so in the place of tan x tan tan 3x what we can replace tan 3x - tan 2x - tan x integral tan 3x dxus integral tan 2x dxus integral tan x dx what is tan x >> log mod x here 3x is d so 1 by 3 log mod >> 3x here 1 by 2 log mod secant 2xus log x 1x 3x = 3 dx = b dx = 1 by 3 dt then it is integral 1 by 3 tan t dt.

[00:18:12]So 1x3 log mod seeant t same Foster jail.

[00:20:05]All right.

[00:20:51]Good.

[00:20:59]There.

[00:21:16]X by A + B tan² X integral tan X by A + B tan² Yes.

[00:21:50]Integral tan x by a + b tan² x It is also have to simplify integral tan x by a + b sin^ 2x by cos^ 2x X.

[00:22:36]Okay. So if you take LCM here, this is A into cos² X. Take cos square X. Then the denominator cos square transpose numerator tan x means sin x by into cos² x by >> a cos² x + >> b sin square x dx.

[00:23:06]Okay. Here a cos² x + b sin² x is equal to p substitution.

[00:23:21]These are all the methods in integration by substitution.

[00:23:25]So a into cos square x into cos x2 plus b into sin square x2 >> sin x2 into dx is equal to.

[00:23:47]So what we can take out from all this >> sin 2 sin x cos x mean sin 2x >> sin 2x into >> b - a dx = dt. So here it is cos² cos cancel here what we require.

[00:24:11]>> Here only sin x cos x is there.

[00:24:14]So 1 by 2 into >> sin 2x by a cos² x + >> b sin² x dx. So this is sin 2x dx is = 1x b - a dt.

[00:24:39]So sin 2x dx numerator 1 by 2 * b - a integral dt by is equal to 1 by 2 into b - a >> log mod t + c. So 1x 2 * b - a log mod t who is d a cos² x >> plus b sin² x c Integral sin x + a by sin x + b dx Next question. Integral sin x + a by sin x + b dx.

[00:28:05]Somewhere question.

[00:28:20]Huh?

[00:28:29]So integral sin x + a by x + b. So here denominator should be sin t or sin x something. So let x + b = t x = t - b dx = d. Then we can simplify integral sin x x t - b + a by sin t dx=.

[00:29:12]So integral sin t cos a minus b plus cos t >> sin a minus b by >> sin t then you can make it sin t sin t gets cancel cos a minus b * integral 1 d 1 dt plus ru sin a minus b has come as a constant separate cos t by sin b.

[00:29:54]So this is t cos a - b into b plus sin a minus b into log mod sin d which formula Huh. Integral fash by >> f_sub_x dx that is equal to >> log fx dative cos. So fd of x by so substitute the value t here cos a - b * x + b + sin a - b * mod sin x + There's What's going on?

[00:34:24]Chemistry and physics classes in Peru.

[00:34:40]Right.

[00:34:45]Integral 1x² + a² dx 1x² + a² dx.

[00:35:00]It is direct formula 1 >> 1 by a tan >> 1 by a >> tan inverse >> x by a. So how you will get this?

[00:35:19]Let x = tan a tan theta or x = a tan theta or a co theta whatever then you can simplify it h x= a Huh?

[00:35:50]So what is dx?

[00:35:54]A into sec square theta d theta integral 1 by a² + a² tan² theta a² + x² a² tan square theta into dx a square theta d theta right so here a square has common 1x a² square integral 1x 1 + tan square theta into a sec square theta.

[00:36:33]So see square gets cancel a 1 cancel 1x a integral 1 d theta that is equal to 1 by a into theta that is equal to 1x into what is theta from here >> tan inverse >> x that's the required integral 1 by x² -² dx X.

[00:37:04]Next one. Listen. Integral 1 by x² - a² dx chemistry physics that is equal to integral 1 by this is x - a into x + a dx like sin x - a sin x - b in the same way if you take here integral x + a - x - a by x - a * x + a dx x - x gets cancel a minus of minus plus a there is no 2 a in numerator so by 2 Right? Then what happened here? 1x 2 a * integral x + a x + a gets cancel 1 by x - a dxus x - x - gets cancel 1 by >> x + a d x that is equal to 1x dt 1x 2 a into log x - a - log x + a. So that is equal to 1x 2 a into log x - a by x +. So 1x integral 1 by x² - a² dx = 1 by 2 a log x - a by x Let's see.

[00:39:22]Integral 1 by a² - x² dx integral 1 by a² - x² >> dx. So 1 by a + x into - x >> a - x >> dx >> dx.

[00:39:45]So integral a + x plus a - x by a + x into a - x 1x 2 a dx.

[00:40:01]So 1x 2 a integral a + x a + x gets cancel 1 by >> dx plus integral 1 by >> a + x >> dx >> dxicient x - x - dx = dt then dx = >> minus so 1 by 2 a * x log x + ah otherwise okay dx plus minus dt minus integral 1 by t dt plus 1x 1x u du a + x = u dx = d a + = right= 1 by 2 a * this is log u minus log right so 1 by 2 a log mod u a + x by >> a so 1x a square - x² dx 1 by 2 a log mod a + x by okay What's Hey, you Okay.

[00:47:28]Next one.

[00:47:34]Integral Integral 1 by<unk> / x² - a² dx is equal to log x +<unk> / a² x² - a² Okay.

[00:48:21]Here x = a secant theta.

[00:48:37]Then dx = a into secant theta tan theta d theta. dx = theta tan theta d theta integral dx a sec theta tan theta d theta upon roo<unk> over x² what is x seeant a seeant so roo<unk> / a² is in Comma see square theta - 1 that is equal to integral what happened here see square - 1 >> tan square roo<unk> tan square tan so a sec theta tan theta by a tan theta d theta so integral theta d theta M* theta + tan theta in both numerator and denominator.

[00:50:00]Now these square then this is integral fdash of x by fx is equal to log mod fx plus c.

[00:50:21]So this is log mod seeant theta + tan theta.

[00:50:28]So that is equal to log uh seeant theta. What is seeant theta integral? Second theta x by >> a bigger than x by a plus what about tan theta x² - a² chain Okay.

[00:51:15]square 1 by a roo<unk> / x² - a² root x² - only tan theta x² x +<unk> / x² - a² by a log mus.

[00:52:02]So log m x +<unk> / x² - a² modus conant log a this is minus log a conant Okay.

[00:52:32]Integral 1x a² + x² roo<unk> / dx is equal to log x +<unk> / x² + a² or a square + x² Here put x = a tan theta.

[00:53:10]Same x = a tan theta.

[00:58:19]Hello.

[00:58:21]Uh this one here uh a = x = a cos 2 theta x = a cos 2 theta x = a cos 2 integral roo<unk> / x - a by b - x dx R integral roo<unk> / x - a into x - b dx = a cos² alpha plus b sin² beta A cos² theta + b sin square theta.

[01:00:38]So sorry for six.

[01:00:45]Hm.

[01:00:52]Formula 6 is equal to 1 by a tan inverse.

[01:01:01]>> What is formula 7?

[01:01:08]>> A squareus is x² - a square. Okay.

[01:01:16]1x 2 a log x - a >> x + integral 1 by a² - x² dx 1x 2 a + x by a 1x roo<unk> / x² - >> a² Log x +<unk> / x² - a² integral 1 by<unk> / a² + x² >> log mod x + roo<unk> / >> a² + integral 1 by<unk> / >> a²us >> sin inverse X by A.

[01:02:16]So these six results are going to be used in which concept?

[01:02:24]First one is integral 1 by a X² + Bx + C dx.

[01:02:42]And the second one is integral 1 by<unk> / a x² + bx + c dx. In these two types these any of these six have to use right. So first of all, a x² + b x + c into complete squaring form.

[01:03:32]Complete squaring form.

[01:03:34]a x² + b x + c into complete squaring form.

[01:03:49]is that a x² + b x + c a into x² + b by a x + c by a into x² + 2 into x into bx 2 a + b by 2 a square - b / 2 a² + c by a a 2 into 2x 2 a b square a square b square add square huh so a into this is x + b by 2 a square + c by a - b² square a square x 2 a² + 4 a c - b² by 4 a square* with a a into x + b by 2 a² + 4 a c - b² by 4 a square 1 cancel 4 a this is = a x² + b x plus C.

[01:05:26]So, for example, Integral 1 by 2x² - x + 1 dx plus x - right.

[01:06:09]First of all complete squaring with here 2x² + x - 1 whoever x² coefficient whatever taken as common >> x² + >> 2 into x² + 2 into x into 1x 4 nothing but xx2 into 2x2 here >> - 1 by 2. So 2 into x² + 2 into x into 1x 4 + b >> b² 1x 16 - 1x 16 - 1 by >> 2 into x + 1x 4² - LCM 16 9 by 16 and 10 16 1's 2 8 into 1 - 1 - 8 - 9 right so we do 1 by 2 integral 1 by x + 1x 4² - 3x 4² dx second for 1x 2 into 1 by 2 a 3 by into log mod x - a x + 1 by 4 - by >> x + 1x 4 + modulus Yes.

[01:08:10]Oh no.

[01:08:19]1 by 3 log mod x - >> 1 by >> x + 1.

[01:08:38]Simplify change 1 by 3 log mod 2x - 1 by 2 into that's 10 important concept Tell me I And I feel a product.

[01:14:23]Okay.

[01:14:53]Mumbai.

[01:15:23]Next.

[01:15:28]Sorry.

[01:15:57]10.

[01:16:04]Example, integral cos x by integral cos x by<unk> /<unk> / sin² x - 2 sin x - 3 So this is the question.

[01:16:56]So first of all I have to remove the sin x. Simply we can remove H. So like sin x is equal to t cos x dx = dt. So numerator is that cos x dx dt integral 1 by<unk> / t²us 2dus 3d complete squaring so t² - 2t - 3 t² - 2 into t into 1 okay + 1² - 1 square - 3 so this part becomes t - 1 squareus 2² right so integral 1 by<unk> / >> t -1² - 2² >> d so which one >> fourth one x²us >> a² integral 1x<unk> / x² - a² log mod x + roo<unk> / x² - a square. So this is log mod x + t -1 roo<unk> / x² - a². So, x² next tus. So t - 1 square - a square and 2 >> 2 log mod what is t >> sin x so sin x -1 +<unk> / sin² x - 2 sin x - 3 yes that's the required Okay, good.

[01:20:41]Integral roo<unk> / secant x - 1 dx 1 by 1 integral 1 by<unk> / 1 - e^ Snakes integral<unk> / X by AQ - XQ DX X integral roo<unk> over sin X - alpha by sin X + A H complete the four questions.

[01:23:43]A come.

[01:24:14]Integral 1 by<unk> / 1 - e^ 2x X that is equal to integral 1 by<unk> / e 2x as common e - 2x -1 dx 2x into - 2x 1x right 2x e X numerator by roo<unk> / e - x² -1 dx e 2x roo<unk> e x - x e - 2x - x² right e - x - e power - x dx = dt e - x dx only minus dt minus dtus integral dt by<unk> / e - x t² - 1 which formula t² - 1 >> x² - a² root both - log mod t + roo<unk> / t² - 1 + c that is equal to minus log who T e power - x +<unk> / t² - 1 e - 2x - 1 + 10 Teaspoon.

[01:29:59]Third one integral <unk>x by<unk> / aq minus x^ 3x2 whole square dx.

[01:30:32]Listen root x separate the numerator root for numerator denominator.

[01:30:41]This x can be written as x 3x2 square root.

[01:30:50]Okay.

[01:30:52]x 3x2= what is it?

[01:30:57]3x2 x^ 3x2 -1 dx = d2 right so x^ 3x2 - 1 by 2 rootx dx= 2x 3 dt so 2x3 integral dt by<unk> / a power 3x2²us - >> t² formula roo<unk> a square - x square sin inverse >> x by a >> x by a so 2x 3 sin inverse x t by a a 3x. So what is t? x power >> that is equal to >> 2x 3 sin inverse x by a whole power.

[01:32:31]roo<unk> x - 1 that is also same roo<unk> 1 - cos x by cos x dx inte x 1 + cos x by 1 + cos x dx. So integral sin x by<unk> cos² x + cos x dx 1 - sin square 1 - cos² sin square roo<unk> sin square sin cos x = t that cos x = A - sin x dx = dt sin x dx dt integral minus dt by<unk> / t² + t minus integral dt by<unk> / t² + 2 into t into 1 by 2 + 1x 2² - 1 by 2² that is equal to minus integral d by this is uh t + 1 by 2²us 1x2 square root Right.

[01:34:54]Huh?

[01:34:58]X² - >> A square then X² - A square roo<unk> X² - A square and fourth result log X + roo<unk> X square - >> A square so - log X + T +<unk> T² - a² 1 by 2.

[01:35:30]So t integral t + 1 by 2 t + 1 by 2 + roo<unk> / t² cos x chair.

[01:35:52]Sorry.

[01:39:56]Right.

[01:39:58]So next question. Integral<unk> / sin x - alpha by x + d x that is equal to integral roo<unk> / rationalize the denominator sin x - ala by rationalize the numerator sin x + into sin x - alpha by sin x - alpha rationalize the numerator uh denominator.

[01:40:35]Okay. Then what happened here? Numerator sin x - alpha square under root becomes sin x - alpha sin x cos alpha minus cos x >> sin >> by what about denominator? sin a + b into sin a minus b<unk> sin square x - >> alpha sin square x - okay dx now cos alpha integral sin x by<unk> / this is 1 - cos² this is 1 - cos² so becomes cos² alpha - cos² x dxus constant sin alpha integral cos x by<unk> / sin² x - sin² alpha X, right? So now cos X = T - sin X DX = DV.

[01:42:28]sin x = t cos x dx = so minus gravity minus cos alpha. So sin x dx dt by<unk> / cos alpha² - t² then u - sin alpha integral d by roo<unk> / t² - sin alpha square gray formula usually x² - a square under roo<unk>2 a square - x square under <unk>2 x² - a square under <unk>2 a square - x² under<unk>2 so first one a squareus this one second one this one is a squareus sin inverse X by A - sin alpha into log modu +<unk> / sin square. So what is t sin x here subur right?

[01:44:38]So complete next model.

[01:44:49]Okay. Okay.

[01:45:13]So evening 5 to6 physics 6 to7 chemistry