# Current Affairs JEE Main & Advanced

#### Triangle Theorem of Couples

If three forces acting on a body be represented in magnitude, direction and line of action by the sides of triangle taken in order, then they are equivalent to a couple whose moment is represented by twice the area of triangle.     Consider the force P along AE, Q along CA and R along AB. These forces are three concurrent forces acting at A and represented in magnitude and direction by the sides BC, CA and AB of DABC. So, by the triangle law of forces, they are in equilibrium.     The remaining two forces P along AD and P along BC form a couple, whose moment is, $m=P.AL=BC.AL$    Since, $\frac{1}{2}(BC.AL)$ = Area of the $\Delta ABC$   $\therefore$Moment $=BC.AL=2$(Area of $\Delta ABC$).

#### Integration By Substitution

(1) When integrand is a function i.e., $\int{f\mathbf{[}\varphi \mathbf{(}x\mathbf{)}]\,\varphi '\mathbf{(}x\mathbf{)}\,dx}$:     Here, we put $\varphi (x)=t,$ so that $\varphi '(x)dx=dt$ and in that case the integrand is reduced to $\int{f(t)dt}$.     (2) When integrand is the product of two factors such that one is the derivative of the others i.e., $I=\int{f(x).{f}'(x).dx}$: In this case we put $f(x)=t$ and convert it into a standard integral.     (3) Integral of a function of the form $f\mathbf{(}ax+b\mathbf{)}$: Here we put $ax+b=t$ and convert it into standard integral. Obviously if $\int{f(x)dx=\varphi (x),}$ then $\int{f(ax+b)dx=\frac{1}{a}\varphi (ax+b)}+c$.     (4) If integral of a function of the form $\frac{{f}'(x)}{f(x)}$   $\int_{{}}^{{}}{\frac{{f}'\text{(}x\text{)}}{f\text{(}x\text{)}}\,}dx=\log \,[f(x)]+c$     (5) If integral of a function of the form ${{[f(x)]}^{n}}{f}'(x)$ $\int{{{[f\text{(}x\text{)}]}^{n}}{f}'\text{(}x\text{)}\,dx=\frac{{{[f\text{(}x\text{)}]}^{n+1}}}{n+1}}+c$,    $\text{ }\!\![\!\!\text{ }n\ne -1\text{ }\!\!]\!\!\text{ }$     (6) If the integral of a function of the form $\frac{{f}'(x)}{\sqrt{f(x)}}$ $\int{\frac{{f}'\text{(}x\text{)}}{\sqrt{f\text{(}x\text{)}}}dx\,=2\sqrt{f\text{(}x\text{)}}+c}$     (7) Standard substitutions
Integrand form Substitution
(i) $\sqrt{{{a}^{2}}-{{x}^{2}}},\frac{1}{\sqrt{{{a}^{2}}-{{x}^{2}}}},{{a}^{2}}-{{x}^{2}}$ $x=a\sin \theta ,$ or $x=a\cos \theta$
(ii) $\sqrt{{{x}^{2}}+{{a}^{2}}},\,\frac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}},\,{{x}^{2}}+{{a}^{2}}$ $x=a\tan \theta$ or $x=a\sin \text{h}\theta$
(iii) $\sqrt{{{x}^{2}}-{{a}^{2}},}\,\,\frac{1}{\sqrt{{{x}^{2}}-{{a}^{2}}}},\,\,{{x}^{2}}-{{a}^{2}}$ $x=a\sec \theta$ or $x=a\cosh \theta$
(iv) $\sqrt{\frac{x}{a+x}},\,\,\sqrt{\frac{a+x}{x}},\,\,\sqrt{x(a+x)},\,\,\frac{1}{\sqrt{x(a+x)}}$ $x=a{{\tan }^{2}}\theta$
more...

#### Couples

Two equal unlike parallel forces which do not have the same line of action, are said to form a couple.     Example : Couples have to be applied in order to wind a watch, to drive a gimlet, to push a cork screw in a cork or to draw circles by means of  pair of compasses.     (1) Arm of the couple : The perpendicular distance between the lines of action of the forces forming the couple is known as the arm of the couple.     (2) Moment of couple : The moment of a couple is obtained in magnitude by multiplying the magnitude of one of the forces forming the couple and perpendicular distance between the lines of action of the force. The perpendicular distance between the forces is called the arm of the couple. The moment of the couple is regarded as positive or negative according as it has a tendency to turn the body in the anticlockwise or clockwise direction.     Moment of a couple = Force $\times$ Arm of the couple = P.p     (3) Sign of the moment of a couple : The moment of a couple is taken with positive or negative sign according as it has a tendency to turn the body in the anticlockwise or clockwise direction.
• A couple can not be balanced by a single force, but can be balanced by a couple of opposite sign.

#### Moment

The moment of a force about a point O is given in magnitude by the product of the forces and the perpendicular distance of O from the line of action of the force.     If F be a force acting at point A of a rigid body along the line AB and $OM(=p)$  be the perpendicular distance of the fixed point O from AB, then the moment of force about O     $=F.p=AB\times OM=2\left[ \frac{1}{2}(AB\times OM) \right]$$=2\text{ }(\text{Area of }\Delta AOB)$      The S.I. unit of moment is Newton-meter $(N-m)$.     (1) Sign of the moment : The moment of a force about a point measures the tendency of the force to cause rotation about that point. The tendency of the force ${{F}_{1}}$ is to turn the lamina in the clockwise direction and of the force ${{F}_{2}}$ is in the anticlockwise direction.     The usual convention is to regard the moment which is anticlockwise direction as positive and that in the clockwise direction as negative.     (2) Varignon's theorem : The algebraic sum of the moments of any two coplanar forces about any point in their plane is equal to the moment of their resultant about the same point.
• The algebraic sum of the moments of any two forces about any point on the line of action of their resultant is zero.

• Conversely, if the algebraic sum of the moments of any two coplanar forces, which are not in equilibrium, about any point in their plane is zero, their resultant passes through the point.

• If a body, having one point fixed, is acted upon by two forces and is at rest. Then the moments of the two forces about the fixed point are equal and opposite.

#### Parallel Forces

(1) Like parallel forces : Two parallel forces are said to be like parallel forces when they act in the same direction.     The resultant $R$ of two like parallel forces $P$ and $Q$ is equal in magnitude of the sum of the magnitudes of forces and $R$ acts in the same direction as the forces $P$ and $Q$ and at the point on the line segment joining the point of action $P$ and $Q,$ which divides it in the ratio $Q:P$ internally.     (2) Two unlike parallel forces : Two parallel forces are said to be unlike if they act in opposite directions.     If $P$ and $Q$ be two unlike parallel forces acting at $A$ and $B$ and $P$ is greater in magnitude than $Q$. Then their resultant $R$ acts in the same direction as $P$ and acts at a  point $C$ on $BA$ produced. Such that $R=P-Q$ and $P.CA=Q.CB$     Then in this case $C$ divides $BA$ externally in the inverse ratio of the forces, $\frac{P}{CB}=\frac{Q}{CA}=\frac{P-Q}{CB-CA}=\frac{R}{AB}$.

#### Lami's Theorem

If three forces acting at a point be in equilibrium, each force is proportional to the sine of the angle between the other two. Thus if the forces are $P,\,\,Q$ and $R;$ $\alpha ,\,\,\beta ,\,\,\gamma$  be the angles between $Q$ and $R,\,\,R$ and $P,\,\,P$ and $Q$ respectively, also the forces are in equilibrium, we have, $\frac{P}{\sin \alpha }=\frac{Q}{\sin \beta }=\frac{R}{\sin \gamma }$.         The converse of this theorem is also true.

#### Polygon Law of Forces

If any number of forces acting on a particle be represented in magnitude and direction by the sides of a polygon taken in order, the forces shall be in equilibrium.

#### Triangle Law of Forces

If three forces, acting at a point, be represented in magnitude and direction by the sides of a triangle, taken in order, they will be in equilibrium.     Here $\overrightarrow{AB}=P,\ \ \ \overrightarrow{BC}=Q,\ \ \overrightarrow{CA}=R$     In triangle ABC, we have $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=0$     $\Rightarrow P+Q+R=0$     Hence the forces $P,\,\,Q,\,\,R$ are in equilibrium.     Converse : If three forces acting at a point are in equilibrium, then they can be represented in magnitude and direction by the sides of a triangle, taken in order.

#### Parallelogram Law of Forces

If two forces, acting at a point, be represented in magnitude and direction by the two sides of a parallelogram drawn from one of its angular points, their resultant is represented both in magnitude and direction of the parallelogram drawn through that point.     If $OA$ and $OB$ represent the forces $P$ and $Q$ acting at a point $O$ and inclined to each other at an angle$\alpha$. If $R$ is the resultant of these forces represented by the diagonal $OC$ of the parallelogram $OACB$ and $R$ makes an angle $\theta$  with $P$.     i.e., $\angle COA=\theta$, then ${{R}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ\cos \alpha$ and $\tan \theta =\frac{Q\sin \alpha }{P+Q\cos \alpha }$     The angle ${{\theta }_{1}}$ which the resultant $R$ makes with the direction of the force $Q$ is given by ${{\theta }_{1}}={{\tan }^{-1}}\left( \frac{P\sin \alpha }{Q+P\cos \alpha } \right)$     Case (i) : If $P=Q$     $\therefore R=2P\cos \,\left( \frac{\alpha }{2} \right)$ and $\tan \theta =\tan \left( \frac{\alpha }{2} \right)$ or $\theta =\frac{\alpha }{2}$       Case (ii) : If $\alpha =90{}^\circ$, i.e. forces are perpendicular     $\therefore R=\sqrt{{{P}^{2}}+{{Q}^{2}}}$and $\tan \theta =\frac{Q}{P}$     Case (iii) : If $\alpha =0{}^\circ$, i.e. forces act in the same direction     $\therefore {{R}_{\max }}=P+Q$     Case (iv) : If $\alpha =180{}^\circ$, i.e. forces act in opposite direction     $\therefore {{R}_{\min }}=P-Q$
• The resultant of two forces is closer to the larger force.

• The resultant of two equal forces of magnitude $P$ acting at an angle $\alpha$ is $2P\cos \frac{\alpha }{2}$ and it bisects the angle between the forces.

• If the resultant R of two forces $P$ and $Q$ acting at an angle $\alpha$ makes an angle $\theta$  with the direction of $P,$ then  $\sin \theta =\frac{Q\sin \alpha }{R}$and $\cos \theta =\frac{P+Q\cos \alpha }{R}$

• If the resultant $R$ of the forces $P$ and $Q$ acting at an angle $\alpha$ makes an angle $\theta$ with the direction of the force $Q,$ then $\sin \theta =\frac{P\sin \alpha }{R}$ and $\cos \theta =\frac{Q+P\sin \alpha }{R}$
Component of a force in two directions : The component of a force $R$ in two directions making angles $\alpha$  and $\beta$  with the line of action of $R$ on and opposite sides of it are         ${{F}_{1}}=\frac{OC.\sin \beta }{\sin (\alpha +\beta )}=\frac{R\sin \beta }{\sin (\alpha +\beta )}$     and          ${{F}_{2}}=\frac{OC.\sin \alpha }{\sin (\alpha +\beta )}=\frac{R.\sin \alpha }{\sin (\alpha +\beta )}$     $\mathbf{\lambda r\mu }$ theorem : The resultant of two forces acting at a point $O$ in directions $OA$ and $OB$represented in magnitudes by $\lambda .OA$and $\mu .OB$ respectively is represented more...

#### Fundamental Integration Formulae

(1) (i) $\int{{{x}^{n}}dx=\frac{{{x}^{n+1}}}{n+1}+c,\,n\ne -1}$     (ii) $\int{{{(ax+b)}^{n}}dx=\frac{1}{a}.\,\frac{{{(ax+b)}^{n+1}}}{n+1}}+c$, $n\ne -1$     (2) (i) $\int{\frac{1}{x}dx=\log |x|+c}$     (ii) $\int{\frac{1}{ax+b}\,dx=\frac{1}{a}(\log |ax+b|)+c}$     (3) $\int{{{e}^{x}}dx={{e}^{x}}+c}$           (4)  $\int{{{a}^{x}}dx=\frac{{{a}^{x}}}{{{\log }_{e}}a}+c}$       (5)  $\int{\sin x\,dx=-\cos x+c}$               (6)  $\int{\cos x\,dx=\sin x+c}$                (7)  $\int{{{\sec }^{2}}}x\,dx=\tan x+c$                   (8)  $\int{\text{cos}\text{e}{{\text{c}}^{2}}x\,dx}=-\cot x+c$                 (9) $\int{\sec x\,\tan x\,dx=\sec x+c}$                    (10) $\int{\text{cosec}\,x\,\cot x\,dx=-\text{cosec}\,x+c}$     (11) $\int{\tan x\,dx}=-\log |\cos x|+c=\log |\sec x|+c$     (12) $\int{\cot x\,dx=\log |\sin x|+}c=-\log |\cos ec\,x|+c$                 (13) $\int{\sec x\,dx=\log |\sec x+\tan x|+c=\log \tan \left( \frac{\pi }{4}+\frac{x}{2} \right)+c}$     (14) $\int{\text{cos}\text{ec}}\,x\,dx=\log |\text{cos}\text{ec}\,x-\cot x|+c=\log \tan \frac{x}{2}+c$     (15) $\int{\frac{dx}{\sqrt{1-{{x}^{2}}}}={{\sin }^{-1}}x+c=-{{\cos }^{-1}}x+c}$       (16) $\int{\frac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}={{\sin }^{-1}}\frac{x}{a}+c}=-{{\cos }^{-1}}\frac{x}{a}+c$     (17) $\int{\frac{dx}{1+{{x}^{2}}}={{\tan }^{-1}}x+c=-{{\cot }^{-1}}x+c}$     (18) $\int{\frac{dx}{{{a}^{2}}+{{x}^{2}}}=\frac{1}{a}{{\tan }^{-1}}\frac{x}{a}+c=\frac{-1}{a}{{\cot }^{-1}}\frac{x}{a}+c}$     (19) $\int{\frac{dx}{x\sqrt{{{x}^{2}}-1}}={{\sec }^{-1}}x+c}=-\cos e{{c}^{-1}}x+c$                 (20) $\int{\frac{dx}{x\sqrt{{{x}^{2}}-{{a}^{2}}}}=\frac{1}{a}{{\sec }^{-1}}\frac{x}{a}+c}=\frac{-1}{a}\cos e{{c}^{-1}}\frac{x}{a}+c$     In any of the fundamental integration formulae, if $x$ is replaced by $ax+b$, then the same formulae is applicable but we must divide by coefficient of $x$ or derivative of $(ax+b)$ i.e., $a$. In general, if $\int{f(x)dx=\varphi (x)+c}$, then$\int{f(ax+b)\,dx=\frac{1}{a}\varphi \,(ax+b)}+c$     $\int{\sin (ax+b)\,dx=\frac{-1}{a}}\cos (ax+b)+c,$     $\int{\sec (ax+b)\,dx=\frac{1}{a}\log |\sec (ax+b)+\tan (ax+b)|+c}$ etc.     Some more results :     (i) $\int{\frac{1}{{{x}^{2}}-{{a}^{2}}}=\frac{1}{2a}\log \left| \frac{x-a}{x+a} \right|+c=\frac{-1}{a}{{\coth }^{-1}}\frac{x}{a}+c}$, when $x>a$     (ii) $\int{\frac{1}{{{a}^{2}}-{{x}^{2}}}dx=\frac{1}{2a}\log \left| \frac{a+x}{a-x} \right|+c=\frac{1}{a}{{\tanh }^{-1}}\frac{x}{a}+c}$, when $x<a$     (iii) $\int{\frac{dx}{\sqrt{{{x}^{2}}-{{a}^{2}}}}=\log \{|x+\sqrt{{{x}^{2}}-{{a}^{2}}}|\}+c=\cos \,{{\text{h}}^{-1}}\left( \frac{x}{a} \right)}+c$     (iv) $\int{\frac{dx}{\sqrt{{{x}^{2}}+{{a}^{2}}}}=\log }\{|x+\sqrt{{{x}^{2}}+{{a}^{2}}}|\}+c=\sin {{\text{h}}^{-1}}\left( \frac{x}{a} \right)+c$     (v) $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\frac{1}{2}x\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{1}{2}{{a}^{2}}{{\sin }^{-1}}\left( \frac{x}{a} \right)+c}$     (vi) $\int_{{}}^{{}}{\sqrt{{{x}^{2}}-{{a}^{2}}}}dx=\frac{1}{2}x\sqrt{{{x}^{2}}-{{a}^{2}}}-\frac{1}{2}{{a}^{2}}\log \{x+\sqrt{{{x}^{2}}-{{a}^{2}}}\}+c$$=\frac{1}{2}x\sqrt{{{x}^{2}}-{{a}^{2}}}-\frac{1}{2}{{a}^{2}}{{\cosh }^{-1}}\left( \frac{x}{a} \right)+c$     (vii) $\int{\sqrt{{{x}^{2}}+{{a}^{2}}}dx=\frac{1}{2}x\sqrt{{{x}^{2}}+{{a}^{2}}}+\frac{1}{2}{{a}^{2}}\log \{x+\sqrt{{{x}^{2}}+{{a}^{2}}}\}+c}$$=\frac{1}{2}x\sqrt{{{x}^{2}}+{{a}^{2}}}+\frac{1}{2}{{a}^{2}}\sin {{\text{h}}^{-1}}\left( \frac{x}{a} \right)$

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