JEE Main & Advanced

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 \[\Delta ABC\]).  

    (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   more...

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 more...

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 more...

(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 more...

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.

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.      

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.  

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 \[P=Q\] more...

(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) \[\int{\frac{dx}{x\sqrt{{{x}^{2}}-{{a}^{2}}}}=\frac{1}{a}{{\sec }^{-1}}\frac{x}{a}+c}=\frac{-1}{a}\cos e{{c}^{-1}}\frac{x}{a}+c\]     more...